elementary superfunctions

[Kouznetsov]

It should be noted that superfunction is not unique in most cases. For example, for
\( f(x)=2 x^2-1 \), superfunction is \( F(x)=\cos(2^x C) \)

Ya, this is the simple kind of non-uniqueness, its just a translation along the x-axis.
However there are also more severe types of non-uniques, as I already introduced in my first post, we have two solutions (which are not translations of each other):
\( F(x)=\cos(2^x) \) and \( F(x)=\cosh(2^x) \).

1. Sorry, Henryk, they are translations of each other.
\( \cos(2^x)=\cosh\Big(2^{x+\pi\cdot i\cdot \ln(2)/2}\Big) \).
We already had similar discussion with respect to tetration on base \( \sqrt{2} \),
http://www.ils.uec.ac.jp/~dima/PAPERS/2009sqrt2.pdf , figure 3.
the growing up SuperExponential (red) and the tetration (blue), at the appropriate translations formula (5.7) and formula (5. become very similar and bounded along the real axis functions (green). (I do not know why the number of forumla that follows (5.7) becomes some strange "smile". I mean just the number of formula, nothing more.)

2. There are many ways to extend the table of superfunctions.
I suggest the group of transforms of the pairs (TransferFunciton, SuperFunctions).

<b>Theorem</b>.
Let \( f(F(z)=F(z+1) \),
Let \( q(p(z))=z \),
Let \( h(z)=p(h(q(z))) \),
Let \( E(z)=p(F(z)) \).
Then \( h(E(z))=E(z+1) \).
Proof:
\( h(E(z))=p(f(q(p(F(z)))))=p(f(F(z)))=p(F(z+1))=E(z+1) \)
(end of proof).

With transform \( p \), from the pair (f,F) we get the pair (h,E).

2.1. Also, in the right hand side of the expression \( f(z)=F(1+F^{-1}(z)) \)
we can swap \( F \) and \( F^{-1} \);
this gives the new transfer function \( h(z)=F^{-1}(1+F(z)) \)
with known superfunction \( F^{-1} \).