> 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.

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5 hours ago | parent | prev | next [-]

[deleted]

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hyperhello 6 hours ago | parent | prev [-]

eml(1,eml(x,1)) = eml(eml(1,x),1) = exp(ln(x)) = ln(exp(x)) = x

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thaumasiotes 4 hours ago | parent | next [-]

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?

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freehorse 2 hours ago | parent | prev [-]

eml(1,eml(x,1)) = e + x

and

eml(eml(1,x),1) = e^e * x

	▲	hyperhello 3 minutes ago \| parent [-]
		Okay, I’m tired. Not quite inverse but per the title , must be a way.