← Back to context Comment by hyperhello 15 hours ago eml(1,eml(x,1)) = eml(eml(1,x),1) = exp(ln(x)) = ln(exp(x)) = x 4 comments hyperhello Reply thaumasiotes 13 hours ago 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? freehorse 11 hours ago eml(1,eml(x,1)) = e + xandeml(eml(1,x),1) = e^e * x hyperhello 9 hours ago Okay, I’m tired. Not quite inverse but per the title , must be a way. 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
thaumasiotes 13 hours ago 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?
freehorse 11 hours ago eml(1,eml(x,1)) = e + xandeml(eml(1,x),1) = e^e * x hyperhello 9 hours ago Okay, I’m tired. Not quite inverse but per the title , must be a way. 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
hyperhello 9 hours ago Okay, I’m tired. Not quite inverse but per the title , must be a way. 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
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?
eml(1,eml(x,1)) = e + x
and
eml(eml(1,x),1) = e^e * x
Okay, I’m tired. Not quite inverse but per the title , must be a way.
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