Ok I want to talk about the connection between superfunction operator , left-distributive and analytic continuation.
First the superfunction is not unique but computing what a function is the superfunction of , is almost unique ; just a single parameter usually.
If we have a function f(x,y) that is analytic in x and y and we take the superfunction F(z,x,y) (with the same method) WITH respect to x , for every fixed y , where z is the number of iterations of f(x,y) ( with respect to x ) then F(z,x,y) is usually analytic in both x and y !
Therefore the superfunction operator is an analytic operator.
this makes going from x <s> y to x <s+1> y - for sufficiently large s - preserve analyticity.
Secondly we want
x <s> 1 = x for all s.
by doing that we set going from x <s> y to x <s+1> y as a superfunction operator.
This gives us an opportunity to get analytic hyperoperators.
Combining x <s> 1 = x , the superfunction method going from x <s> y to x <s+1> y and the left distributive property to go from going from x <s> y to x <s-1> y we then get a nice structure for hyperoperators that connects to the ideas of iterations and superfunctions.
You see we then get that x <s> y is EXACTLY the y th iterate of x < s-1> y with respect to x and starting value y. If we set y = 1 then x <s> 1 = x thereby proving that it is indeed taking superfunctions *we start with x * (for all s).
This implies that
x <0> y = x + y is WRONG.
We get by the above :
x < 0 > y = x + y - 1
( x <0> 1 = x !! )
x < 1 > y = x y
( the super of +x + 1 - 1 aka +x y times )
x < 2 > y = x^y
( the super of x y ; taking x * ... y times )
x < 3 > y = x^^ y
( starting at x and repeating x^... )
This also allows us to compute x < n > y for any n , even negative.
That is a sketch of my idea.
Not sure how this relates to 2 < s > 2 = 4 ...
Now we only need to understand x < s > y for s between 0 and 1 but analytic at 0 and 1.
Gotta run.
Regards
tommy1729
Tom Marcel Raes
First the superfunction is not unique but computing what a function is the superfunction of , is almost unique ; just a single parameter usually.
If we have a function f(x,y) that is analytic in x and y and we take the superfunction F(z,x,y) (with the same method) WITH respect to x , for every fixed y , where z is the number of iterations of f(x,y) ( with respect to x ) then F(z,x,y) is usually analytic in both x and y !
Therefore the superfunction operator is an analytic operator.
this makes going from x <s> y to x <s+1> y - for sufficiently large s - preserve analyticity.
Secondly we want
x <s> 1 = x for all s.
by doing that we set going from x <s> y to x <s+1> y as a superfunction operator.
This gives us an opportunity to get analytic hyperoperators.
Combining x <s> 1 = x , the superfunction method going from x <s> y to x <s+1> y and the left distributive property to go from going from x <s> y to x <s-1> y we then get a nice structure for hyperoperators that connects to the ideas of iterations and superfunctions.
You see we then get that x <s> y is EXACTLY the y th iterate of x < s-1> y with respect to x and starting value y. If we set y = 1 then x <s> 1 = x thereby proving that it is indeed taking superfunctions *we start with x * (for all s).
This implies that
x <0> y = x + y is WRONG.
We get by the above :
x < 0 > y = x + y - 1
( x <0> 1 = x !! )
x < 1 > y = x y
( the super of +x + 1 - 1 aka +x y times )
x < 2 > y = x^y
( the super of x y ; taking x * ... y times )
x < 3 > y = x^^ y
( starting at x and repeating x^... )
This also allows us to compute x < n > y for any n , even negative.
That is a sketch of my idea.
Not sure how this relates to 2 < s > 2 = 4 ...
Now we only need to understand x < s > y for s between 0 and 1 but analytic at 0 and 1.
Gotta run.
Regards
tommy1729
Tom Marcel Raes