> eml(x,y)=exp(x)-ln(y)
Exp and ln, isn't the operation its own inverse depending on the parameter? What a neat find.
> 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?
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?