Exponential Mean and Beyond

[tommy1729]

What about using commutative hyper-operations to do means? Then the means would already be commutative.

true. 

but why do you want a new average ? 
what is the purpose ?

this reminds me of the matrix question 

is A^B equal to exp(log(A) B) or exp(B log(A))

and both are defendable.

regards

tommy1729