if perhaps I read you wrong, and maybe you meant to say
\( (f^{\diamond -n} \diamond f^{\diamond n})(x)= f(x) \)
\( (f^{\diamond -n} \diamond f^{\diamond n+1}(g^{\diamond n + 1}(x) + 1))(x) = f^{\diamond 1}(g^{\diamond n + 1}(x) + 1) \)
this would imply inconsistency.
Which to me is no surprise, the diamond operator was just suggested in a makeshift attempt to come up with a recurrence relation for fractional values of superfunction indexes. It was by no means canon.
It hardly makes the rest of my post "inconsistent" as if you read carefully I even asked [in reference to the diamond operator]:
That being said, the concept of
\( g^{\diamond n}(f^{\diamond n}(x)) = x \)
\( f^{\diamond n}(x) = f^{\diamond n+1}(g^{\diamond n+1}(x) + 1) \)
is still perfectly consistent which was the main point of the post. And it is hardly wrong at first glance.
therefore I still can't help but think by your post that you mistook \( \diamond \) for \( \circ \)
\( (f^{\diamond -n} \diamond f^{\diamond n})(x)= f(x) \)
\( (f^{\diamond -n} \diamond f^{\diamond n+1}(g^{\diamond n + 1}(x) + 1))(x) = f^{\diamond 1}(g^{\diamond n + 1}(x) + 1) \)
this would imply inconsistency.
Which to me is no surprise, the diamond operator was just suggested in a makeshift attempt to come up with a recurrence relation for fractional values of superfunction indexes. It was by no means canon.
It hardly makes the rest of my post "inconsistent" as if you read carefully I even asked [in reference to the diamond operator]:
Quote:So I'm wondering, is this just absurd? It's really the best I can come up with...
That being said, the concept of
\( g^{\diamond n}(f^{\diamond n}(x)) = x \)
\( f^{\diamond n}(x) = f^{\diamond n+1}(g^{\diamond n+1}(x) + 1) \)
is still perfectly consistent which was the main point of the post. And it is hardly wrong at first glance.
therefore I still can't help but think by your post that you mistook \( \diamond \) for \( \circ \)