4 ms·
> 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?
by thaumasiotes 7mo 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?
- hyperhello 7mo agoeml(1,eml(x,1)) = eml(eml(1,x),1) = exp(ln(x)) = ln(exp(x)) = x
- thaumasiotes 7mo agoBut 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?
- freehorse 7mo agoeml(1,eml(x,1)) = e + x and eml(eml(1,x),1) = e^e * x
- hyperhello 7mo agoOkay, I’m tired. Not quite inverse but per the title , must be a way.
- freehorse 7mo agoI was mistaken above in the first identity, it is eml(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
- deleted 7mo ago[deleted]
- woopsn 7mo agoIt'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.