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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replyeml(1,eml(x,1)) = eml(eml(1,x),1) = exp(ln(x)) = ln(exp(x)) = x
replyBut 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?
replyeml(1,eml(x,1)) = e + x
replyand
eml(eml(1,x),1) = e^e * x
Okay, I’m tired. Not quite inverse but per the title , must be a way.
replyI was mistaken above in the first identity, it is
replyeml(1,eml(x,1)) = e - x
Which then if you iterate gives x (ie is inverse of itself).
eml(1,eml(eml(1,eml(x,1)),1)) = x