alt Hacker News> isn't the operation its own inverse depending on the parameter?
This is a function from ℝ² to ℝ. It can't be its own inverse; what would that mean?
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hyperhello • today at 4:28 AM
eml(1,eml(x,1)) = eml(eml(1,x),1) = exp(ln(x)) = ln(exp(x)) = x
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It's a kind of superposition representation a la Kolmogorov-Arnold, a learnable functional basis for elementary functions g(x,y)=f(x) - f^{-1}(y) in this sense with f=exp.