Comment by hyperhello 9 hours ago Okay, I’m tired. Not quite inverse but per the title , must be a way. 1 comment hyperhello Reply freehorse 8 hours ago I was mistaken above in the first identity, it iseml(1,eml(x,1)) = e - xWhich then if you iterate gives x (ie is inverse of itself).eml(1,eml(eml(1,eml(x,1)),1)) = x
freehorse 8 hours ago I was mistaken above in the first identity, it iseml(1,eml(x,1)) = e - xWhich then if you iterate gives x (ie is inverse of itself).eml(1,eml(eml(1,eml(x,1)),1)) = x
I 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