10/23/2007, 11:07 AM
Well, you write:
e#x*e#y , which means (e#x)*(e#y) and you are right because bracketing is not necessary, due to the (hyper)operation's prìorities, and then if you mean (which is not exactly what you wrote):
e#x*e#y = e#(x*y) as a clone of e^x*e^y = e^(x+y) unfortunately, it's wrong
(e#x)#y = e#(x^y) as a clone of (e^x)^y = e^(x*y) it is also wrong
(e#x)#((x^e)#y) as a clone of (e^x)^(e^y) = e^((x*e)^y)) ... idem .... !!
No luck! This is the question. The problem is much more complicated. The hyper-exponents of tetration don't follow the same or similar rules of the lower rank (hyper)-operations.
No problem, we shall overcome!
GFR
e#x*e#y , which means (e#x)*(e#y) and you are right because bracketing is not necessary, due to the (hyper)operation's prìorities, and then if you mean (which is not exactly what you wrote):
e#x*e#y = e#(x*y) as a clone of e^x*e^y = e^(x+y) unfortunately, it's wrong
(e#x)#y = e#(x^y) as a clone of (e^x)^y = e^(x*y) it is also wrong
(e#x)#((x^e)#y) as a clone of (e^x)^(e^y) = e^((x*e)^y)) ... idem .... !!
No luck! This is the question. The problem is much more complicated. The hyper-exponents of tetration don't follow the same or similar rules of the lower rank (hyper)-operations.
No problem, we shall overcome!
GFR