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Every inverse function is its own graph, reflected across y = x. The inverse trig functions have a “restricted domain” because you can’t mirror a curve that doubles back. The faded arcs are what we discard.

The Math Flow

86,660 subscribers

25,501 Aufrufe • vor 1 Monat •via X (Twitter)

Anya Rossi• Live Now

Private livecam show

5 Kommentare

Profilbild von even constant

even constantvor 1 Monat

arctan is my favorite

Profilbild von The Math Flow

The Math Flowvor 1 Monat

The reason being?

Profilbild von Varaha mihir

Varaha mihirvor 1 Monat

Here's an interesting video i found

Profilbild von ryan laff

ryan laffvor 1 Monat

Nice visualizations

Profilbild von mukeshchaurasia

mukeshchaurasiavor 1 Monat

@grok explain in simpler terms

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This man teaches at a community college in California. His salary: around $800,000 a year. The engineers who passed calculus because of him: $1,800,000 to start. He has more calculus students than Harvard, MIT, and Stanford combined. This is Professor Leonard's Calculus 2, Lecture 6.2. Free on YouTube. Professor Leonard has taught calculus on YouTube for over a decade. His channel has millions of subscribers across 150 countries. Every major university has students who watch him the night before their exam. Then the concept. An inverse function is a machine that undoes another machine. If a function takes 2 and gives you 8, the inverse takes 8 and gives you back 2. Finding an inverse means switching every x and y in the equation and solving for y again. The graph flips across the line y = x like a mirror. Then the problem. Sometimes it is easy to find the inverse. Sometimes it is impossible to write it explicitly. A function like 3π sin x + sin x cannot be solved for x with algebra. You have to think. What angle makes the whole thing equal to 1? You work backwards through the unit circle until the answer appears. Then the shortcut. If you want the derivative of an inverse at a point, you do not need the inverse itself. You only need the derivative of the original function. The formula: the derivative of the inverse at a point equals 1 divided by the derivative of the original function evaluated at the switched point. The inverse flips the coordinates, so you flip where you plug in. Watch the moment he shows why G prime of 8 equals 1/12 without ever writing the inverse function. Every engineering student memorizes the derivative rules. Professor Leonard's lecture is the one that shows why the inverse derivative formula is just those same rules run backwards. A software engineer at a semiconductor company in Austin said Professor Leonard's channel is the reason she passed Calculus 2 on her second attempt. She graduated, joined the company, and now makes $165,000 a year. Bookmark this and watch later - after this lecture every inverse problem on your exam will feel like a question you already answered.

This man teaches at a community college in California. His salary: around $800,000 a year. The engineers who passed calculus because of him: $1,800,000 to start. He has more calculus students than Harvard, MIT, and Stanford combined. This is Professor Leonard's Calculus 2, Lecture 6.2. Free on YouTube. Professor Leonard has taught calculus on YouTube for over a decade. His channel has millions of subscribers across 150 countries. Every major university has students who watch him the night before their exam. Then the concept. An inverse function is a machine that undoes another machine. If a function takes 2 and gives you 8, the inverse takes 8 and gives you back 2. Finding an inverse means switching every x and y in the equation and solving for y again. The graph flips across the line y = x like a mirror. Then the problem. Sometimes it is easy to find the inverse. Sometimes it is impossible to write it explicitly. A function like 3π sin x + sin x cannot be solved for x with algebra. You have to think. What angle makes the whole thing equal to 1? You work backwards through the unit circle until the answer appears. Then the shortcut. If you want the derivative of an inverse at a point, you do not need the inverse itself. You only need the derivative of the original function. The formula: the derivative of the inverse at a point equals 1 divided by the derivative of the original function evaluated at the switched point. The inverse flips the coordinates, so you flip where you plug in. Watch the moment he shows why G prime of 8 equals 1/12 without ever writing the inverse function. Every engineering student memorizes the derivative rules. Professor Leonard's lecture is the one that shows why the inverse derivative formula is just those same rules run backwards. A software engineer at a semiconductor company in Austin said Professor Leonard's channel is the reason she passed Calculus 2 on her second attempt. She graduated, joined the company, and now makes $165,000 a year. Bookmark this and watch later - after this lecture every inverse problem on your exam will feel like a question you already answered.

Lupen

736,457 Aufrufe • vor 1 Monat

2. LAMBDA Use a LAMBDA function to create custom, reusable functions and call them by a friendly name. This is a powerful function to create your own calculations. In our case, we wanted to calculate the income after deducting tax. The formula: = LAMBDA(x,y,x*(1-y))

2. LAMBDA Use a LAMBDA function to create custom, reusable functions and call them by a friendly name. This is a powerful function to create your own calculations. In our case, we wanted to calculate the income after deducting tax. The formula: = LAMBDA(x,y,x*(1-y))

Kofi Ntim🇬🇭

11,082 Aufrufe • vor 1 Jahr

The 'inverse Cramer' thing is the best engagement hack for a content creator personality. Because it just means that anything you say, you just get a million impressions from people saying the opposite.

The 'inverse Cramer' thing is the best engagement hack for a content creator personality. Because it just means that anything you say, you just get a million impressions from people saying the opposite.

TBPN

47,544 Aufrufe • vor 6 Monaten

Do you know why it's so hard to detect incoming missiles and decoys? Because it falls into a class problem called Inverse Problems. Most people are far more familiar with forward problems. You know the system, you know the inputs, and you work out what happens next. Inverse Problems go the other way. You see messy measurements and from them you have to reconstruct the hidden thing you actually care about. That is a much harder task. A Radar does not see a missile at x with velocity v. It sees noisy returns, clutter, ghost echoes, and dropouts, and from that it has to infer whats actually out there.

Do you know why it's so hard to detect incoming missiles and decoys? Because it falls into a class problem called Inverse Problems. Most people are far more familiar with forward problems. You know the system, you know the inputs, and you work out what happens next. Inverse Problems go the other way. You see messy measurements and from them you have to reconstruct the hidden thing you actually care about. That is a much harder task. A Radar does not see a missile at x with velocity v. It sees noisy returns, clutter, ghost echoes, and dropouts, and from that it has to infer whats actually out there.

Mathelirium

83,018 Aufrufe • vor 7 Monaten

Drop everything and listen in to 9.45 minutes of Leader of Opposition Rahul Gandhi ji’s speech delivered at the INDIA bloc meeting on June 8th “The Congress Party is a resistance movement protecting the idea that all Indians are equal. If political parties can’t function, what functions? Resistance functions. Resistance works. Wherever we resist, it works. I have seen it with my own eyes. I have walked 4,000 kilometres across this country - resistance works. You don't need political architecture. You don't need the bureaucracy. You don't need the intelligence agencies. You need the act of resistance — meaning: I will resist. I will not allow injustice. Full stop, the end. It is a spirit. It is not an organisation. It is a way of thinking - that is where we have to go.”

Drop everything and listen in to 9.45 minutes of Leader of Opposition Rahul Gandhi ji’s speech delivered at the INDIA bloc meeting on June 8th “The Congress Party is a resistance movement protecting the idea that all Indians are equal. If political parties can’t function, what functions? Resistance functions. Resistance works. Wherever we resist, it works. I have seen it with my own eyes. I have walked 4,000 kilometres across this country - resistance works. You don't need political architecture. You don't need the bureaucracy. You don't need the intelligence agencies. You need the act of resistance — meaning: I will resist. I will not allow injustice. Full stop, the end. It is a spirit. It is not an organisation. It is a way of thinking - that is where we have to go.”

Supriya Shrinate

51,456 Aufrufe • vor 3 Monaten

For a country that hates its own, I think this is the worst kind of analysis you can have about what we produce because you paint a picture that we don't produce anything good. It doesn't help anyone.

For a country that hates its own, I think this is the worst kind of analysis you can have about what we produce because you paint a picture that we don't produce anything good. It doesn't help anyone.

98Patrobas🇰🇪

148,401 Aufrufe • vor 2 Monaten

Ask anyone who’s taken a course in Ordinary Differential Equations (ODEs) what a solution to an ODE represents geometrically, and most of them won’t have a clean answer. When I first took ordinary differential equations, the pattern was always the same. Early on it turns into a speedrun of methods: separation of variables, integrating factors, variation of parameters, Bernoulli, exact equations. Then pretty quickly the course slides into hammer-picking. Spot the form, apply the recipe, move on. Too mechanical! And the real problem is what you don’t walk away with. You leave with a toolkit, but without a feel for what a differential equation even is, especially geometrically. That matters because in real modeling the equations you meet are rarely nice enough to reward memorised recipes. So you get trained to solve toy forms, while the actual subject stays blurry. The behavior. The flow. The shape of solutions. It wasn't until I watched the first lecture of Professor Arthur Mattuck that I realized I didn’t actually know what a solution to a differential equation represents geometrically. His point is almost embarrassingly simple. A first-order ODE is a slope field, and a solution is a curve that stays tangent to that field everywhere. The math breakdown: Write the ODE as dy/dx = f(x,y). At each point (x,y), attach a tiny line segment with slope f(x,y). A function y = y₁(x) is a solution exactly when its graph follows those slopes. At every x, the slope of the curve equals the slope prescribed by the field at the point on the curve. That’s the one line that ties both viewpoints together: y₁′(x) = f(x, y₁(x)). So solving the ODE and drawing an integral curve are the same statement in two languages. Once you see that, you stop obsessing over whether you can write y(x) in closed form. You start asking the questions that actually matter. Where do solutions flow. Where do they get trapped. Where do they blow up. Where does existence or uniqueness fail because the field isn’t even defined? That’s the perspective shift I wish every ODE course forces early. It’s also why I keep pairing math with animation. #DifferentialEquations #ODEs #VectorFields #AppliedMathematics #Mathematics #

Ask anyone who’s taken a course in Ordinary Differential Equations (ODEs) what a solution to an ODE represents geometrically, and most of them won’t have a clean answer. When I first took ordinary differential equations, the pattern was always the same. Early on it turns into a speedrun of methods: separation of variables, integrating factors, variation of parameters, Bernoulli, exact equations. Then pretty quickly the course slides into hammer-picking. Spot the form, apply the recipe, move on. Too mechanical! And the real problem is what you don’t walk away with. You leave with a toolkit, but without a feel for what a differential equation even is, especially geometrically. That matters because in real modeling the equations you meet are rarely nice enough to reward memorised recipes. So you get trained to solve toy forms, while the actual subject stays blurry. The behavior. The flow. The shape of solutions. It wasn't until I watched the first lecture of Professor Arthur Mattuck that I realized I didn’t actually know what a solution to a differential equation represents geometrically. His point is almost embarrassingly simple. A first-order ODE is a slope field, and a solution is a curve that stays tangent to that field everywhere. The math breakdown: Write the ODE as dy/dx = f(x,y). At each point (x,y), attach a tiny line segment with slope f(x,y). A function y = y₁(x) is a solution exactly when its graph follows those slopes. At every x, the slope of the curve equals the slope prescribed by the field at the point on the curve. That’s the one line that ties both viewpoints together: y₁′(x) = f(x, y₁(x)). So solving the ODE and drawing an integral curve are the same statement in two languages. Once you see that, you stop obsessing over whether you can write y(x) in closed form. You start asking the questions that actually matter. Where do solutions flow. Where do they get trapped. Where do they blow up. Where does existence or uniqueness fail because the field isn’t even defined? That’s the perspective shift I wish every ODE course forces early. It’s also why I keep pairing math with animation. #DifferentialEquations #ODEs #VectorFields #AppliedMathematics #Mathematics #

Mathelirium

40,841 Aufrufe • vor 7 Monaten

Found a setup that replaces a team of researchers with 300 AI agents and one shared graph. the interesting part is what happens after all 300 come back with different pieces of the answer. because now you don’t have 300 agents. you have: 300 outputs → hundreds of sources → repeated entities → overlapping claims → contradictions → thousands of possible relationships and all of that has to collapse into one structure. that’s what the context graph is doing. instead of: agent → report you get: agents → sources → entities → edges → contradictions → verification → one graph every source becomes a node. every shared entity creates a connection. 300 agents can create information. the graph is what turns that information into memory the next agent can actually use. that’s the anatomy behind the setup ↓

Found a setup that replaces a team of researchers with 300 AI agents and one shared graph. the interesting part is what happens after all 300 come back with different pieces of the answer. because now you don’t have 300 agents. you have: 300 outputs → hundreds of sources → repeated entities → overlapping claims → contradictions → thousands of possible relationships and all of that has to collapse into one structure. that’s what the context graph is doing. instead of: agent → report you get: agents → sources → entities → edges → contradictions → verification → one graph every source becomes a node. every shared entity creates a connection. 300 agents can create information. the graph is what turns that information into memory the next agent can actually use. that’s the anatomy behind the setup ↓

Ricker

19,652 Aufrufe • vor 27 Tagen

What if Your Neural Network Was Forced to Obey Physics? Physics-Informed Neural Networks (PINNs) are neural networks trained to satisfy a differential equation by building the PDE residual directly into the loss. They emerged from a very practical problem...classical PDE pipelines can be brilliant, but they often demand heavy discretization work (meshes, stencils, stability tuning), and the method you build is usually tied to one geometry and one solver setup. A PINN flips the workflow by representing the solution itself as a smooth function uᵩ(x,t) and enforcing the physics everywhere you choose to sample the domain. People often meet PINNs in the least helpful way...via a flashy solution plot, and almost no explanation of what was enforced to get it. In this series we keep the enforcement visible. We pick a differential equation, represent the unknown solution as a flexible function, measure how well that function satisfies the equation across the domain, and train it to reduce that mismatch everywhere we sample. A normal neural net learns from labels...you give it inputs and target outputs. A PINN learns from a differential equation...you give it inputs (x,t) and it gets punished whenever its output fails the PDE. By punish we mean that the loss increases when the mismatch is large we reward it if the loss decreases as the mismatch gets smaller. The network isn’t replacing physics, it’s becoming a flexible function that is forced to satisfy the same calculus you’d impose on any candidate solution. The math breakdown: We start with a PDE we want to solve on a domain Ω. Write it as uₜ(x,t) + N(u(x,t), uₓ(x,t), uₓₓ(x,t), …) = 0 for (x,t) in Ω A PINN replaces the unknown function u with a neural network output uᵩ(x,t) Now define the physics residual by plugging uᵩ into the PDE rᵩ(x,t) = ∂uᵩ/∂t + N(uᵩ, ∂uᵩ/∂x, ∂²uᵩ/∂x², …) If uᵩ were an exact solution, we would have rᵩ(x,t) = 0 everywhere. We may also have data points (xᵢ,tᵢ,uᵢ) from measurements or a known initial condition. The training objective is just a weighted sum of squared errors L(ᵩ) = L_data(ᵩ) + λ L_phys(ᵩ) + L_bc/ic(ᵩ) with L_data(ᵩ) = meanᵢ |uᵩ(xᵢ,tᵢ) − uᵢ|² L_phys(ᵩ) = meanⱼ |rᵩ(xⱼ,tⱼ)|² where (xⱼ,tⱼ) are the collocation points in Ω L_bc/ic(ᵩ) = penalties enforcing boundary conditions and initial conditions The key technical step is that the derivatives inside rᵩ are computed by automatic differentiation ∂uᵩ/∂t, ∂uᵩ/∂x, ∂²uᵩ/∂x², … So we can differentiate the total loss L(ᵩ) with respect to ᵩ and train with gradient descent. This is the whole idea behind PINNs. Learn a function, but make the PDE part of the loss, so the network is trained to be a solution, not just a curve-fitter. In the render, the main 3D surface is the network’s current guess uᵩ(x,t), drawn as a living sheet over the (x,t) plane. Hovering above is the neural scaffold...a visible graph of feature nodes and connections. The bright tension threads are the physics residual rᵩ(x,t): each thread tethers a collocation bead on the sheet up to the scaffold, and it thickens and brightens exactly where |rᵩ| is large (color encodes the sign). As training runs, those threads go slack across the domain not because we hid the error, but because the network has actually been pushed toward rᵩ(x,t) ≈ 0. #PINNs #PhysicsInformedNeuralNetworks #ScientificMachineLearning #PDE #DifferentialEquations #Optimization #MachineLearning #AppliedMath #ComputationalPhysics

What if Your Neural Network Was Forced to Obey Physics? Physics-Informed Neural Networks (PINNs) are neural networks trained to satisfy a differential equation by building the PDE residual directly into the loss. They emerged from a very practical problem...classical PDE pipelines can be brilliant, but they often demand heavy discretization work (meshes, stencils, stability tuning), and the method you build is usually tied to one geometry and one solver setup. A PINN flips the workflow by representing the solution itself as a smooth function uᵩ(x,t) and enforcing the physics everywhere you choose to sample the domain. People often meet PINNs in the least helpful way...via a flashy solution plot, and almost no explanation of what was enforced to get it. In this series we keep the enforcement visible. We pick a differential equation, represent the unknown solution as a flexible function, measure how well that function satisfies the equation across the domain, and train it to reduce that mismatch everywhere we sample. A normal neural net learns from labels...you give it inputs and target outputs. A PINN learns from a differential equation...you give it inputs (x,t) and it gets punished whenever its output fails the PDE. By punish we mean that the loss increases when the mismatch is large we reward it if the loss decreases as the mismatch gets smaller. The network isn’t replacing physics, it’s becoming a flexible function that is forced to satisfy the same calculus you’d impose on any candidate solution. The math breakdown: We start with a PDE we want to solve on a domain Ω. Write it as uₜ(x,t) + N(u(x,t), uₓ(x,t), uₓₓ(x,t), …) = 0 for (x,t) in Ω A PINN replaces the unknown function u with a neural network output uᵩ(x,t) Now define the physics residual by plugging uᵩ into the PDE rᵩ(x,t) = ∂uᵩ/∂t + N(uᵩ, ∂uᵩ/∂x, ∂²uᵩ/∂x², …) If uᵩ were an exact solution, we would have rᵩ(x,t) = 0 everywhere. We may also have data points (xᵢ,tᵢ,uᵢ) from measurements or a known initial condition. The training objective is just a weighted sum of squared errors L(ᵩ) = L_data(ᵩ) + λ L_phys(ᵩ) + L_bc/ic(ᵩ) with L_data(ᵩ) = meanᵢ |uᵩ(xᵢ,tᵢ) − uᵢ|² L_phys(ᵩ) = meanⱼ |rᵩ(xⱼ,tⱼ)|² where (xⱼ,tⱼ) are the collocation points in Ω L_bc/ic(ᵩ) = penalties enforcing boundary conditions and initial conditions The key technical step is that the derivatives inside rᵩ are computed by automatic differentiation ∂uᵩ/∂t, ∂uᵩ/∂x, ∂²uᵩ/∂x², … So we can differentiate the total loss L(ᵩ) with respect to ᵩ and train with gradient descent. This is the whole idea behind PINNs. Learn a function, but make the PDE part of the loss, so the network is trained to be a solution, not just a curve-fitter. In the render, the main 3D surface is the network’s current guess uᵩ(x,t), drawn as a living sheet over the (x,t) plane. Hovering above is the neural scaffold...a visible graph of feature nodes and connections. The bright tension threads are the physics residual rᵩ(x,t): each thread tethers a collocation bead on the sheet up to the scaffold, and it thickens and brightens exactly where |rᵩ| is large (color encodes the sign). As training runs, those threads go slack across the domain not because we hid the error, but because the network has actually been pushed toward rᵩ(x,t) ≈ 0. #PINNs #PhysicsInformedNeuralNetworks #ScientificMachineLearning #PDE #DifferentialEquations #Optimization #MachineLearning #AppliedMath #ComputationalPhysics

Mathelirium

17,459 Aufrufe • vor 4 Monaten

Lecture 1 on Physics-Informed Neural Networks: A Mini-Series Physics-Informed Neural Networks (PINNs) are neural networks trained to satisfy a differential equation by building the PDE residual directly into the loss. They emerged from a very practical problem...classical PDE pipelines can be brilliant, but they often demand heavy discretization work (meshes, stencils, stability tuning), and the method you build is usually tied to one geometry and one solver setup. A PINN flips the workflow by representing the solution itself as a smooth function uᵩ(x,t) and enforcing the physics everywhere you choose to sample the domain. People often meet PINNs in the least helpful way...via a flashy solution plot, and almost no explanation of what was enforced to get it. In this series we keep the enforcement visible. We pick a differential equation, represent the unknown solution as a flexible function, measure how well that function satisfies the equation across the domain, and train it to reduce that mismatch everywhere we sample. A normal neural net learns from labels...you give it inputs and target outputs. A PINN learns from a differential equation...you give it inputs (x,t) and it gets punished whenever its output fails the PDE. By punish we mean that the loss increases when the mismatch is large we reward it if the loss decreases as the mismatch gets smaller. The network isn’t replacing physics, it’s becoming a flexible function that is forced to satisfy the same calculus you’d impose on any candidate solution. The math breakdown: We start with a PDE we want to solve on a domain Ω. Write it as uₜ(x,t) + N(u(x,t), uₓ(x,t), uₓₓ(x,t), …) = 0 for (x,t) in Ω A PINN replaces the unknown function u with a neural network output uᵩ(x,t) Now define the physics residual by plugging uᵩ into the PDE rᵩ(x,t) = ∂uᵩ/∂t + N(uᵩ, ∂uᵩ/∂x, ∂²uᵩ/∂x², …) If uᵩ were an exact solution, we would have rᵩ(x,t) = 0 everywhere. We may also have data points (xᵢ,tᵢ,uᵢ) from measurements or a known initial condition. The training objective is just a weighted sum of squared errors L(ᵩ) = L_data(ᵩ) + λ L_phys(ᵩ) + L_bc/ic(ᵩ) with L_data(ᵩ) = meanᵢ |uᵩ(xᵢ,tᵢ) − uᵢ|² L_phys(ᵩ) = meanⱼ |rᵩ(xⱼ,tⱼ)|² where (xⱼ,tⱼ) are the collocation points in Ω L_bc/ic(ᵩ) = penalties enforcing boundary conditions and initial conditions The key technical step is that the derivatives inside rᵩ are computed by automatic differentiation ∂uᵩ/∂t, ∂uᵩ/∂x, ∂²uᵩ/∂x², … So we can differentiate the total loss L(ᵩ) with respect to ᵩ and train with gradient descent. This is the whole idea behind PINNs. Learn a function, but make the PDE part of the loss, so the network is trained to be a solution, not just a curve-fitter. In the render, the main 3D surface is the network’s current guess uᵩ(x,t), drawn as a living sheet over the (x,t) plane. Hovering above is the neural scaffold...a visible graph of feature nodes and connections. The bright tension threads are the physics residual rᵩ(x,t): each thread tethers a collocation bead on the sheet up to the scaffold, and it thickens and brightens exactly where |rᵩ| is large (color encodes the sign). As training runs, those threads go slack across the domain not because we hid the error, but because the network has actually been pushed toward rᵩ(x,t) ≈ 0. #PINNs #PhysicsInformedNeuralNetworks #ScientificMachineLearning #PDE #DifferentialEquations #Optimization #MachineLearning #AppliedMath #ComputationalPhysics

Lecture 1 on Physics-Informed Neural Networks: A Mini-Series Physics-Informed Neural Networks (PINNs) are neural networks trained to satisfy a differential equation by building the PDE residual directly into the loss. They emerged from a very practical problem...classical PDE pipelines can be brilliant, but they often demand heavy discretization work (meshes, stencils, stability tuning), and the method you build is usually tied to one geometry and one solver setup. A PINN flips the workflow by representing the solution itself as a smooth function uᵩ(x,t) and enforcing the physics everywhere you choose to sample the domain. People often meet PINNs in the least helpful way...via a flashy solution plot, and almost no explanation of what was enforced to get it. In this series we keep the enforcement visible. We pick a differential equation, represent the unknown solution as a flexible function, measure how well that function satisfies the equation across the domain, and train it to reduce that mismatch everywhere we sample. A normal neural net learns from labels...you give it inputs and target outputs. A PINN learns from a differential equation...you give it inputs (x,t) and it gets punished whenever its output fails the PDE. By punish we mean that the loss increases when the mismatch is large we reward it if the loss decreases as the mismatch gets smaller. The network isn’t replacing physics, it’s becoming a flexible function that is forced to satisfy the same calculus you’d impose on any candidate solution. The math breakdown: We start with a PDE we want to solve on a domain Ω. Write it as uₜ(x,t) + N(u(x,t), uₓ(x,t), uₓₓ(x,t), …) = 0 for (x,t) in Ω A PINN replaces the unknown function u with a neural network output uᵩ(x,t) Now define the physics residual by plugging uᵩ into the PDE rᵩ(x,t) = ∂uᵩ/∂t + N(uᵩ, ∂uᵩ/∂x, ∂²uᵩ/∂x², …) If uᵩ were an exact solution, we would have rᵩ(x,t) = 0 everywhere. We may also have data points (xᵢ,tᵢ,uᵢ) from measurements or a known initial condition. The training objective is just a weighted sum of squared errors L(ᵩ) = L_data(ᵩ) + λ L_phys(ᵩ) + L_bc/ic(ᵩ) with L_data(ᵩ) = meanᵢ |uᵩ(xᵢ,tᵢ) − uᵢ|² L_phys(ᵩ) = meanⱼ |rᵩ(xⱼ,tⱼ)|² where (xⱼ,tⱼ) are the collocation points in Ω L_bc/ic(ᵩ) = penalties enforcing boundary conditions and initial conditions The key technical step is that the derivatives inside rᵩ are computed by automatic differentiation ∂uᵩ/∂t, ∂uᵩ/∂x, ∂²uᵩ/∂x², … So we can differentiate the total loss L(ᵩ) with respect to ᵩ and train with gradient descent. This is the whole idea behind PINNs. Learn a function, but make the PDE part of the loss, so the network is trained to be a solution, not just a curve-fitter. In the render, the main 3D surface is the network’s current guess uᵩ(x,t), drawn as a living sheet over the (x,t) plane. Hovering above is the neural scaffold...a visible graph of feature nodes and connections. The bright tension threads are the physics residual rᵩ(x,t): each thread tethers a collocation bead on the sheet up to the scaffold, and it thickens and brightens exactly where |rᵩ| is large (color encodes the sign). As training runs, those threads go slack across the domain not because we hid the error, but because the network has actually been pushed toward rᵩ(x,t) ≈ 0. #PINNs #PhysicsInformedNeuralNetworks #ScientificMachineLearning #PDE #DifferentialEquations #Optimization #MachineLearning #AppliedMath #ComputationalPhysics

Mathelirium

47,308 Aufrufe • vor 8 Monaten

What are Physics-Informed Neural Networks (PINNs) Physics-Informed Neural Networks (PINNs) are neural nets trained to satisfy a differential equation. The trick is simple. You bake the PDE residual straight into the loss. They came out of a very practical pain point. Classical PDE pipelines can be amazing, but they often demand a lot of setup work. Meshes. Stencils. Stability tuning. And once you build a solver, it’s usually tied to one geometry and one discretization choice. A PINN flips the workflow. You represent the solution itself as a smooth function uᵩ(x,t) and you enforce the physics wherever you choose to sample the domain. Most people first meet PINNs in the least helpful way. A pretty solution surface, almost no clarity on what was enforced to make it appear. In this series we keep the enforcement visible. We pick a PDE, represent the unknown solution as a flexible function, measure how badly that function violates the equation across the domain, and train it to reduce that mismatch at the points we sample. A normal neural net learns from labels. You give it inputs and target outputs. A PINN learns from an equation. You give it inputs (x,t), and it gets penalized whenever its output fails the PDE. Smaller mismatch means smaller loss. Bigger mismatch means bigger loss. That’s all “punish” and “reward” mean here. The network isn’t replacing physics. It’s just a flexible function that we force to obey the same calculus you’d demand from any candidate solution. The math breakdown: We start with a PDE on a domain Ω. Write it as uₜ(x,t) + N(u(x,t), uₓ(x,t), uₓₓ(x,t), …) = 0 for (x,t) in Ω A PINN replaces the unknown u with a neural network output uᵩ(x,t) Now define the physics residual by plugging uᵩ into the PDE rᵩ(x,t) = ∂uᵩ/∂t + N(uᵩ, ∂uᵩ/∂x, ∂²uᵩ/∂x², …) If uᵩ were an exact solution, we’d have rᵩ(x,t) = 0 everywhere. We may also have data points (xᵢ,tᵢ,uᵢ) from measurements or from an initial condition. The training objective is a weighted sum of squared errors L(ᵩ) = L_data(ᵩ) + λ L_phys(ᵩ) + L_bc/ic(ᵩ) with L_data(ᵩ) = meanᵢ |uᵩ(xᵢ,tᵢ) − uᵢ|² L_phys(ᵩ) = meanⱼ |rᵩ(xⱼ,tⱼ)|² where (xⱼ,tⱼ) are collocation points in Ω L_bc/ic(ᵩ) = penalties enforcing boundary conditions and initial conditions The key technical step is how we get the derivatives inside rᵩ. We don’t approximate them with finite differences. We compute them with automatic differentiation: ∂uᵩ/∂t, ∂uᵩ/∂x, ∂²uᵩ/∂x², … Then we differentiate the total loss L(ᵩ) with respect to ᵩ and train with gradient descent. That’s the whole idea. Learn a function, but make the PDE part of the loss, so the network is trained to be a solution, not just a curve-fitter. In the render, the main 3D surface is the network’s current guess uᵩ(x,t), drawn as a living sheet over the (x,t) plane. Hovering above is the neural scaffold, a visible graph of feature nodes and connections. The bright tension threads are the physics residual rᵩ(x,t). Each thread tethers a collocation bead on the sheet up to the scaffold, and it thickens and brightens exactly where |rᵩ| is large, with color showing the sign. As training runs, those threads go slack across the domain, not because we hid the error, but because the network has actually been pushed toward rᵩ(x,t) ≈ 0. #PINNs #ScientificMachineLearning #PDE #DifferentialEquations #Optimization #MachineLearning #AppliedMath #ComputationalPhysics

What are Physics-Informed Neural Networks (PINNs) Physics-Informed Neural Networks (PINNs) are neural nets trained to satisfy a differential equation. The trick is simple. You bake the PDE residual straight into the loss. They came out of a very practical pain point. Classical PDE pipelines can be amazing, but they often demand a lot of setup work. Meshes. Stencils. Stability tuning. And once you build a solver, it’s usually tied to one geometry and one discretization choice. A PINN flips the workflow. You represent the solution itself as a smooth function uᵩ(x,t) and you enforce the physics wherever you choose to sample the domain. Most people first meet PINNs in the least helpful way. A pretty solution surface, almost no clarity on what was enforced to make it appear. In this series we keep the enforcement visible. We pick a PDE, represent the unknown solution as a flexible function, measure how badly that function violates the equation across the domain, and train it to reduce that mismatch at the points we sample. A normal neural net learns from labels. You give it inputs and target outputs. A PINN learns from an equation. You give it inputs (x,t), and it gets penalized whenever its output fails the PDE. Smaller mismatch means smaller loss. Bigger mismatch means bigger loss. That’s all “punish” and “reward” mean here. The network isn’t replacing physics. It’s just a flexible function that we force to obey the same calculus you’d demand from any candidate solution. The math breakdown: We start with a PDE on a domain Ω. Write it as uₜ(x,t) + N(u(x,t), uₓ(x,t), uₓₓ(x,t), …) = 0 for (x,t) in Ω A PINN replaces the unknown u with a neural network output uᵩ(x,t) Now define the physics residual by plugging uᵩ into the PDE rᵩ(x,t) = ∂uᵩ/∂t + N(uᵩ, ∂uᵩ/∂x, ∂²uᵩ/∂x², …) If uᵩ were an exact solution, we’d have rᵩ(x,t) = 0 everywhere. We may also have data points (xᵢ,tᵢ,uᵢ) from measurements or from an initial condition. The training objective is a weighted sum of squared errors L(ᵩ) = L_data(ᵩ) + λ L_phys(ᵩ) + L_bc/ic(ᵩ) with L_data(ᵩ) = meanᵢ |uᵩ(xᵢ,tᵢ) − uᵢ|² L_phys(ᵩ) = meanⱼ |rᵩ(xⱼ,tⱼ)|² where (xⱼ,tⱼ) are collocation points in Ω L_bc/ic(ᵩ) = penalties enforcing boundary conditions and initial conditions The key technical step is how we get the derivatives inside rᵩ. We don’t approximate them with finite differences. We compute them with automatic differentiation: ∂uᵩ/∂t, ∂uᵩ/∂x, ∂²uᵩ/∂x², … Then we differentiate the total loss L(ᵩ) with respect to ᵩ and train with gradient descent. That’s the whole idea. Learn a function, but make the PDE part of the loss, so the network is trained to be a solution, not just a curve-fitter. In the render, the main 3D surface is the network’s current guess uᵩ(x,t), drawn as a living sheet over the (x,t) plane. Hovering above is the neural scaffold, a visible graph of feature nodes and connections. The bright tension threads are the physics residual rᵩ(x,t). Each thread tethers a collocation bead on the sheet up to the scaffold, and it thickens and brightens exactly where |rᵩ| is large, with color showing the sign. As training runs, those threads go slack across the domain, not because we hid the error, but because the network has actually been pushed toward rᵩ(x,t) ≈ 0. #PINNs #ScientificMachineLearning #PDE #DifferentialEquations #Optimization #MachineLearning #AppliedMath #ComputationalPhysics

Mathelirium

44,924 Aufrufe • vor 8 Monaten

"...the institutions aren't functioning. And you can tell because instead of giving you noisy, good advice, instead of giving you random advice, they give you upside down pyramids. They give you the inverse of the right advice. And that is an alarming fact. It tells you that there is a process in there that knows what to do and tells you the opposite." - Bret Weinstein

"...the institutions aren't functioning. And you can tell because instead of giving you noisy, good advice, instead of giving you random advice, they give you upside down pyramids. They give you the inverse of the right advice. And that is an alarming fact. It tells you that there is a process in there that knows what to do and tells you the opposite." - Bret Weinstein

The DarkHorse Podcast

29,107 Aufrufe • vor 8 Monaten

Lecture 4 on Calculus of Variations You might wonder...If I’m optimizing a shape...a curve, a surface, a whole path, what does "take the derivative and set it to zero" even mean? Do I take the damn derivative with respect to a curve/surface? 🤔 In normal calculus the variable is a number x, so the reflex is clean...f′(x)=0. In calculus of variations the variable is a whole function...the geometry itself, like a curve y(x) (or a surface z(x,y)). So the derivative can’t be a single slope. It has to be a pointwise sensitivity, i.e. how the objective reacts to tiny local deformations. You’re holding a whole shape, like a curve y(x). Your objective isn’t f(x) anymore, it’s a functional J[y], and as we've seen with our first there examples, usually an integral that depends on the entire curve (often through y and y’). To talk about a “derivative”, you do the only thing that makes sense: you nudge the entire curve by a tiny amount and see how J changes. Pick a wiggle shape η(x). It’s not random...it’s any admissible deformation direction. Admissible just means it obeys the constraints. If the endpoints are fixed, you force η(0)=η(1)=0 so the wiggle doesn’t move the endpoints. Then scale that wiggle by a small number ε and define the perturbed curve yε(x)=y(x)+εη(x). Now treat ε like the usual scalar in a Taylor expansion. As ε→0, J[y+εη] expands as J[y+εη] = J[y] + ε · (first-order term depending linearly on η) + o(ε). So the difference is J[y+εη] - J[y] = ε · (linear functional of η) + o(ε). For the standard integral of a Lagrangian problems, that linear functional can be written as an inner product with some function of x: J[y+εη] - J[y] = ε ∫ (δJ/δy)(x) η(x) dx + o(ε). That’s the definition-level meaning of δJ/δy: it’s the unique pointwise sensitivity function that makes this identity true for every admissible η. If δJ/δy is positive at some x, then choosing η negative there decreases J; if δJ/δy is negative there, pushing y upward locally decreases J. It’s literally a map along the curve saying push this way to go downhill. Now translate “set the derivative to zero.” At a minimizer y*, the first-order change must vanish for every admissible wiggle: J[y*+εη] − J[y*] = o(ε) for all η. Plug in the expansion and the ε-term must be zero: ∫ (δJ/δy)(x) η(x) dx = 0 for all admissible η. Here’s the crucial logic step: the only way an integral against every test function η can be zero is if the integrand itself is zero (in the usual sense used in analysis). So you get δJ/δy = 0. For the common case J[y]=∫ L(x, y, y’) dx, you can compute δJ/δy explicitly and it becomes the Euler–Lagrange expression δJ/δy = ∂L/∂y − d/dx(∂L/∂y’). So if you name the Euler–Lagrange residual as “left-hand side” R(x) = ∂L/∂y − d/dx(∂L/∂y’), then “set the derivative to zero” is exactly R(x)=0. That’s why animation works so well. You don’t have to solve R=0 in one shot. You can evolve the curve in an artificial time τ by pushing it in the downhill direction: ∂y/∂τ = −R(y). Where the residual is large, the curve moves a lot; as the residual drains toward zero, the motion dies out and the curve settles into an extremal. In our animations, we start from an intentionally ugly curve/surface. Frame by frame the functional drops, the residual drains away, and the geometry relaxes into an extremal. #CalculusOfVariations #EulerLagrange #FunctionalDerivative #GradientFlow #Optimization #MathAnimation

Lecture 4 on Calculus of Variations You might wonder...If I’m optimizing a shape...a curve, a surface, a whole path, what does "take the derivative and set it to zero" even mean? Do I take the damn derivative with respect to a curve/surface? 🤔 In normal calculus the variable is a number x, so the reflex is clean...f′(x)=0. In calculus of variations the variable is a whole function...the geometry itself, like a curve y(x) (or a surface z(x,y)). So the derivative can’t be a single slope. It has to be a pointwise sensitivity, i.e. how the objective reacts to tiny local deformations. You’re holding a whole shape, like a curve y(x). Your objective isn’t f(x) anymore, it’s a functional J[y], and as we've seen with our first there examples, usually an integral that depends on the entire curve (often through y and y’). To talk about a “derivative”, you do the only thing that makes sense: you nudge the entire curve by a tiny amount and see how J changes. Pick a wiggle shape η(x). It’s not random...it’s any admissible deformation direction. Admissible just means it obeys the constraints. If the endpoints are fixed, you force η(0)=η(1)=0 so the wiggle doesn’t move the endpoints. Then scale that wiggle by a small number ε and define the perturbed curve yε(x)=y(x)+εη(x). Now treat ε like the usual scalar in a Taylor expansion. As ε→0, J[y+εη] expands as J[y+εη] = J[y] + ε · (first-order term depending linearly on η) + o(ε). So the difference is J[y+εη] - J[y] = ε · (linear functional of η) + o(ε). For the standard integral of a Lagrangian problems, that linear functional can be written as an inner product with some function of x: J[y+εη] - J[y] = ε ∫ (δJ/δy)(x) η(x) dx + o(ε). That’s the definition-level meaning of δJ/δy: it’s the unique pointwise sensitivity function that makes this identity true for every admissible η. If δJ/δy is positive at some x, then choosing η negative there decreases J; if δJ/δy is negative there, pushing y upward locally decreases J. It’s literally a map along the curve saying push this way to go downhill. Now translate “set the derivative to zero.” At a minimizer y, the first-order change must vanish for every admissible wiggle: J[y+εη] − J[y*] = o(ε) for all η. Plug in the expansion and the ε-term must be zero: ∫ (δJ/δy)(x) η(x) dx = 0 for all admissible η. Here’s the crucial logic step: the only way an integral against every test function η can be zero is if the integrand itself is zero (in the usual sense used in analysis). So you get δJ/δy = 0. For the common case J[y]=∫ L(x, y, y’) dx, you can compute δJ/δy explicitly and it becomes the Euler–Lagrange expression δJ/δy = ∂L/∂y − d/dx(∂L/∂y’). So if you name the Euler–Lagrange residual as “left-hand side” R(x) = ∂L/∂y − d/dx(∂L/∂y’), then “set the derivative to zero” is exactly R(x)=0. That’s why animation works so well. You don’t have to solve R=0 in one shot. You can evolve the curve in an artificial time τ by pushing it in the downhill direction: ∂y/∂τ = −R(y). Where the residual is large, the curve moves a lot; as the residual drains toward zero, the motion dies out and the curve settles into an extremal. In our animations, we start from an intentionally ugly curve/surface. Frame by frame the functional drops, the residual drains away, and the geometry relaxes into an extremal. #CalculusOfVariations #EulerLagrange #FunctionalDerivative #GradientFlow #Optimization #MathAnimation

Mathelirium

12,186 Aufrufe • vor 9 Monaten

TikTok memecoins are back. "The purist sign possible that we are back, is a tiktoker that everyone knows shilling a token because it’s a viral meme. This is the purist form of bullishness we have seen in months."

TikTok memecoins are back. "The purist sign possible that we are back, is a tiktoker that everyone knows shilling a token because it’s a viral meme. This is the purist form of bullishness we have seen in months."

rasmr

36,524 Aufrufe • vor 8 Monaten

"Every large institution in the world is going to have functions that can be run by AI, functions that can’t be run by AI...There are things that only a doctor should decide, but there are lots of things that an algorithm can decide.” More from Alex Karp:

"Every large institution in the world is going to have functions that can be run by AI, functions that can’t be run by AI...There are things that only a doctor should decide, but there are lots of things that an algorithm can decide.” More from Alex Karp:

Palantir

25,176 Aufrufe • vor 3 Jahren

HE Bobi wine on a video call welcoming back the 6 political prisoners who have been released the machete and others. We are what we are because of you. We are famous because of you. They said.

HE Bobi wine on a video call welcoming back the 6 political prisoners who have been released the machete and others. We are what we are because of you. We are famous because of you. They said.

Brayo🇺🇬

21,467 Aufrufe • vor 2 Monaten

The labs want to rent you intelligence. Palantir is building the stack for you to own it. “We are building a stack that is fully aligned with you capturing and maintaining the alpha of your business. Every single technology decision we are making is aligned with that.“ That is AI sovereignty. We have the product. We have the scale. We know how to do it. Now we take the market.

The labs want to rent you intelligence. Palantir is building the stack for you to own it. “We are building a stack that is fully aligned with you capturing and maintaining the alpha of your business. Every single technology decision we are making is aligned with that.“ That is AI sovereignty. We have the product. We have the scale. We know how to do it. Now we take the market.

Chad Wahlquist

43,085 Aufrufe • vor 2 Monaten

Did you hear what Rachel Reeves said at the end? “reflected in the budget” That means more taxes in the Autumn. All because Labour is utterly failing to grip our welfare bill. We can’t afford Labour.

Did you hear what Rachel Reeves said at the end? “reflected in the budget” That means more taxes in the Autumn. All because Labour is utterly failing to grip our welfare bill. We can’t afford Labour.

Mel Stride MP

41,069 Aufrufe • vor 1 Jahr

A man behind enemy lines, stranded on a mountain, and they still went back for him. Because that’s what the American military does. No matter the odds or the risk, they go. This rescue mission is more than just a success story, it’s a reminder of who we are. We are still a country that fights for its own. Because as long as there are Americans willing to risk everything to bring one of their own home, there is still hope.

A man behind enemy lines, stranded on a mountain, and they still went back for him. Because that’s what the American military does. No matter the odds or the risk, they go. This rescue mission is more than just a success story, it’s a reminder of who we are. We are still a country that fights for its own. Because as long as there are Americans willing to risk everything to bring one of their own home, there is still hope.

Glenn Beck

70,293 Aufrufe • vor 5 Monaten

🔥🚨BREAKING: The creator of the CPU (The brain inside a computer) Federico Faggin has joined the CIA in claiming that reality is a hologram and the WHOLE of reality is contained in its parts, and we are parts CONTAINING THE WHOLE. These claims mirror the CIA’s brain functions report from 1947.

🔥🚨BREAKING: The creator of the CPU (The brain inside a computer) Federico Faggin has joined the CIA in claiming that reality is a hologram and the WHOLE of reality is contained in its parts, and we are parts CONTAINING THE WHOLE. These claims mirror the CIA’s brain functions report from 1947.

Dom Lucre | Breaker of Narratives

666,480 Aufrufe • vor 8 Monaten