The exponential of the derivative translates a function 𝑓(𝑥) by an amount εpic.twitter.com/mwQ1aLpOHt
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We define the exponential of linear operator (i.e. derivative, matrix, etc.) through its Taylor Series The proof then follows directly:https://twitter.com/InertialObservr/status/1145396940122669057?s=20 …
In physics, this is why we call the derivative operator the "generator of translations" pic.twitter.com/iqJoydnylt
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What was the p in the (pc)-squared yesterday?
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Translation is just an epsilon of rotation about the point at infinity!
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Also this is how dual quaternions represent rotation + translation. Unit ℝ quat stores 3D rotation. Epsilon quat stores rotated translation. ε²=0 so product can combine rotations and rotate/add translations without product of translations introducing rotation artifacts.
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Thank you, i hate it. Something tells me there should be a reason why this isn't allowed. Don't people get hurt because of this? You're like Euler :(
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Nothing wrong with passing around operators that have yet to be applied and treating them like numbers... just call them functions and call it lambda calculus and you'll be fine.
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This might be a dumb question. But why is (d/dx)^2 equal to d^2/d(x^2). The numerator is the same, but I can't grasp why the denominator is the same.
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Don't "distribute" the square; (d/dx)^2 f = d/dx (d/dx f) = d^2/dx^2 f
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Thanks. Twitter will use this to make your timeline better. UndoUndo
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