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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.

25,501 次观看 • 1 个月前 •via X (Twitter)

5 条评论

even constant 的头像
even constant1 个月前

arctan is my favorite

The Math Flow 的头像
The Math Flow1 个月前

The reason being?

Varaha mihir 的头像
Varaha mihir1 个月前

Here's an interesting video i found

ryan laff 的头像
ryan laff1 个月前

Nice visualizations

mukeshchaurasia 的头像
mukeshchaurasia1 个月前

@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.

Lupen

736,457 次观看 • 1 个月前

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 次观看 • 7 个月前

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 次观看 • 4 个月前

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 次观看 • 8 个月前

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 次观看 • 8 个月前

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 次观看 • 9 个月前