Superfunction is a multivalued function defined over a set of functions not over a set of numbers:
\( S(f)=F \) means that \( S \) takes a function \( f \) and gives a function \( S(f)=F \) calles superfunction of \( f \)
such that \( F \) satisfies
1) \( F(x+1)=f(F(x)) \)
since there are infinite solution for \( F \) (infinite superfunctions) that means that \( S(f)=F \) is multivalued and then is not a function at all and we have to put some restrictions:
using Trapmann-Kouznetsov terminology used in their paper "5+ methods..." we call \( S_u(f)=F_u \) the \( u \)-based superfunction of \( f \) the function \( F_u(x) \) that satifies two requirements
1) \( F_u(x+1)=f(F_u(x)) \)
2) \( F_u(0)=u \)
and we have
\( F_u(n)=f^{\circ n}(u) \)
In this way we obtain uniqueness over the naturals: in fact superfunction is equivalent to the "definition by recursion" that is unique .
But is not over the reals... there we need more requirments.
Obviously this is still not enough to achieve the uniqueness of \( F_u \) (iteration of \( f \)) that would mean having \( S \) to be a function over a set of functions (not multivalued).
By the way I guess that Trapmann and Kouznetsov tried to find such additionals requirments but my math level is not enough to understand it.
Anyways we have that \( S^{\circ -1} \) is a function and \( S^{\circ 1/2} \) is the half superfunction.
example :
let define \( add_b(x)=b+x \) and \( mul_b(x)=bx \) we have
\( S_0(add_b)=mul_b \) (multiplication is the 0-based superfunction of addition)
so we search for a \( S^{\circ 1/2} \) such that
\( S^{\circ 1/2}(S^{\circ 1/2}(add_b))=mul_b \)
and that if \( b[1,5]x=hyper-(1,5)_b(x) \) we should have (maybe...)
\( S^{\circ 1/2}(add_b)=hyper-(1,5)_b \)
and
\( S^{\circ 1/2}(hyper-(1,5)_b)=mul_b \)
I apologize if I did some mistakes.
\( S(f)=F \) means that \( S \) takes a function \( f \) and gives a function \( S(f)=F \) calles superfunction of \( f \)
such that \( F \) satisfies
1) \( F(x+1)=f(F(x)) \)
since there are infinite solution for \( F \) (infinite superfunctions) that means that \( S(f)=F \) is multivalued and then is not a function at all and we have to put some restrictions:
using Trapmann-Kouznetsov terminology used in their paper "5+ methods..." we call \( S_u(f)=F_u \) the \( u \)-based superfunction of \( f \) the function \( F_u(x) \) that satifies two requirements
1) \( F_u(x+1)=f(F_u(x)) \)
2) \( F_u(0)=u \)
and we have
\( F_u(n)=f^{\circ n}(u) \)
In this way we obtain uniqueness over the naturals: in fact superfunction is equivalent to the "definition by recursion" that is unique .
But is not over the reals... there we need more requirments.
Obviously this is still not enough to achieve the uniqueness of \( F_u \) (iteration of \( f \)) that would mean having \( S \) to be a function over a set of functions (not multivalued).
By the way I guess that Trapmann and Kouznetsov tried to find such additionals requirments but my math level is not enough to understand it.
Anyways we have that \( S^{\circ -1} \) is a function and \( S^{\circ 1/2} \) is the half superfunction.
example :
let define \( add_b(x)=b+x \) and \( mul_b(x)=bx \) we have
\( S_0(add_b)=mul_b \) (multiplication is the 0-based superfunction of addition)
so we search for a \( S^{\circ 1/2} \) such that
\( S^{\circ 1/2}(S^{\circ 1/2}(add_b))=mul_b \)
and that if \( b[1,5]x=hyper-(1,5)_b(x) \) we should have (maybe...)
\( S^{\circ 1/2}(add_b)=hyper-(1,5)_b \)
and
\( S^{\circ 1/2}(hyper-(1,5)_b)=mul_b \)
I apologize if I did some mistakes.
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)
