(11/13/2011, 11:33 PM)JmsNxn Wrote: now I've found a way to generalize this to what I am labeling "meta-superfunctions". These have to do with iteration but in a very different complicated light.
consider the definition:
\( f^{\diamond n}(g^{\diamond n}(x)) = g^{\diamond n}(f^{\diamond n}(x)) = x \)
\( f^{\diamond n}(x) = f^{\diamond n+1}(g^{\diamond n+1}(x) + 1) \)
NO I DONT CONSIDER THE DEFINITION :
the very first equation seems wrong from first glance already ( and im eating chocolate ! )
take the last equation
take \( f^{\diamond -n} \) on both sides
=> \( x = f^{\diamond 1}(g^{\diamond n+1}(x) + 1) \)
( name this equation 3)
this gives the same result as if we had used \( g^{\diamond n} \) on both sides.
hence together with the first equation we conclude f and g are inverse under the diamond operator.
also now consider equation 3 and keep in mind all 3 equations and f and g are inverses then :
take \( g^{\diamond 1} \) on both sides :
=> \( g^{\diamond 1}(x) = (g^{\diamond n+1}(x) + 1) \)
for all n !!!!
this looks really bad.
for small n we get ( n = 0 )
\( g^{\diamond 1}(x) = (g^{\diamond 1}(x) + 1) \)
which seems paradoxal , and actually , it is.
for larger n we get e.g.
\( g^{\diamond 1}(x) = (g^{\diamond 2}(x) + 1) \)
\( g^{\diamond 1}(x) = (g^{\diamond 3}(x) + 1) \)
\( g^{\diamond 1}(x) = (g^{\diamond 4}(x) + 1) \)
\( g^{\diamond 1}(x) = (g^{\diamond 5}(x) + 1) \)
\( g^{\diamond 1}(x) = (g^{\diamond 7}(x) + 1) \)
\( g^{\diamond 1}(x) = (g^{\diamond 1729}(x) + 1) \)
...
till n approaching infinity !!!
simultaneous !
it seems f and g are not functions , the diamond is not a function nor operator nor composition and none of them are finite !?!?
i prefer different definitions and equations.
once again your first post contains mistakes.
all the rest that follows is even more paradoxal.
no offense , but i prefer you look at it first ...
mathematics requires thinking. at least twice.
regards
tommy1729