Comment by thaumasiotes
13 hours ago
> 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?
13 hours ago
> 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?
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.
eml(1,eml(x,1)) = eml(eml(1,x),1) = exp(ln(x)) = ln(exp(x)) = x
But f(x) = eml(1, x) and g(x) = eml(x, 1) are different operations. What operation are you saying is supposed to be its own inverse?
eml(1,eml(x,1)) = e + x
and
eml(eml(1,x),1) = e^e * x
Okay, I’m tired. Not quite inverse but per the title , must be a way.
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