01/11/2015, 02:00 AM
Inspired by the " special addition of velocity in special relativity " I came to consider this a long time ago.
Let x [X] y be the super of x [+] y.
[/] is inverse of [X].
Ex( L(x) [+] L(y) ) = x [X] y.
L is functional inverse of Ex.
Ex(x) = 1 [+] x [+] x [X] x [/] 2! [+] ...
is +,X,/,exp,ln the only solutions ?
regards
tommy1729
Let x [X] y be the super of x [+] y.
[/] is inverse of [X].
Ex( L(x) [+] L(y) ) = x [X] y.
L is functional inverse of Ex.
Ex(x) = 1 [+] x [+] x [X] x [/] 2! [+] ...
is +,X,/,exp,ln the only solutions ?
regards
tommy1729