Some "Theorem" on the generalized superfunction

[JmsNxn]

It took me bout a month revising and getting prepared for my final exam, and finally I made it today

I hope you did well on your exam  

I've been also concentrating myself on the idea of differential forms of an iteration, and it turned out that it is adequate to represent them throughout the iteration velocity, which is also the functional root of Julia's equation.(I'll call it Julia's solution)
So the riemann surface should be established on Julia's equation or its reciprocal. Ramanujan gave his result in his notebook I by
\( \frac{\mathrm{d}f^t(z)}{\mathrm{d}t}=\frac{\mathrm{d}f^t(z)}{\mathrm{d}z}[t^1]f^t(z) \)
where \( [t^1]f^t(z) \) is then proved to be the Julia's solution
we can see that calculating the derivative with respect to t must involve Julia function, itself defined by a differential equation.
(But we have to come to the admission that Julia's equation has infinitely many solutions, correspondingly to the fact that there are infinitely many abel solutions. So the set of all generated Julia's solution should be considered as a category.)
You can see Julia's equation also give a way to describe the infinitesimal iteration and reverse it you have the iteration integral.
Also it proves in some sense the symmetric law, whenever fg=f, then f^t g=f^t(f and f^t treated as branch cuts), because the functional iterations of the same function share the same Julia's solution(by 3rd law, also proveable by many means), for example,
the L denotes Julia's solution
L(cos(z))=L(z)*-sin(z) and L(cos(-z))=L(-z)*sin(z) showing that for some branch cuts L(-z)=L(z)
then by sharing the same L, exchanging cos with arccos
L(arccos(z))=L(z)*-arccos'(z) and L(arccos(-z))=L(-z)*-(arccos(-z))'(z)=-L(-arccos(z))=L(arccos(z))
taking the holomorphic version of inverse of L, arccos(-z)=arccos(z)
which means that in some sense all sets of "-1th-iteration of cos" or arccos(z) has a branch cut arccos(-z) (but normally this is invalid and impractical though)


And I'm concerning if the general composition of two Riemann surface can be represented just as simply the chain rule of d/dx like:
\( \frac{\mathrm{d} h}{\mathrm{d} f}=\frac{\mathrm{d} h}{\mathrm{d} g}\frac{\mathrm{d} g}{\mathrm{d} f} \)
and since f,g,h are three riemann surface, some form of this indeed describe a differential equation for h=g(f) as you may multiply both sides by \( \frac{\mathrm{d} f}{\mathrm{d} x} \)
following the idea that an inverse function's RS is defined this way


Regards, Leo

That is very interesting Leo.

So you want to look at Julia's solution to solve the iterative logarithm problem?

I'm trying to understand exactly what you are trying to do.  Forgive me if I misunderstand you.

To understand the differential equation,

\(
\frac{d}{dt}f^{\circ t}(z) = H(f^{\circ t}(z))\\
\)

Where

\(
H = \frac{d}{dt}|_{t=0}f^{\circ t}(z)\\
\)

We want to use Julia's equation,

\(
F(f(z)) = F(z)f'(z)\\
\)

Where,

\(
\frac{d}{dt}f^{\circ t}(z) = F(z)\frac{d}{dz}f^{\circ t}(z)\\
\)

If I'm interpreting you right; do you mind sourcing me this result? This is absolutely breathtaking.