The exponential of the derivative translates a function 𝑓(𝑥) by an amount εpic.twitter.com/mwQ1aLpOHt
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Mathematical physicist and mentor to mathematically talented youth. Talent is that which bridges the gap between what can be taught and what must be learned.
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The exponential of the derivative translates a function 𝑓(𝑥) by an amount εpic.twitter.com/mwQ1aLpOHt
I tried to interpret this intuitively but am struggling because it relies on the spooky action at a distance enjoyed by analytic functions and is false for general infinite-dimensional functions. Pretty cool though.
that's true .. I think my Physicist is showing .. we damn near always assume analyticity and function/finite number or arguments
No analyticity is needed. As any mathematical physicist knows, there's a self-adjoint operator i d/dx on L^2(R) and a theory of exponentiating self-adjoint operators that makes exp(t d/dx)(f) = f(x+t) true for all functions f in L^2(R). It's not just wishful thinking.
I don't understand this. I can find two infinitely differentiable square-integrable functions that have all the same derivatives at zero but different values at 1. How is that not a counterexample to what you are saying? Or is "as any mathematical physicist knows" ... 1/
Perhaps it will help to note that the mapping U_t on [L^2](R) which sends f(x) to f(x + t) is unitary, and that its infintesimal generator, which sends f(x) to lim[f(x + t) - f(x)]/t, is densely defined, with unique skew-adjoint extension. It is thus "essentially skew-adjoint."
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