Some "Theorem" on the generalized superfunction

[tommy1729]

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

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Finally we compare (4)&(7), we complete the proof, showing that \( \varphi_1(z)=\lambda(z) \), which is also the difinition of the iteration velocity
Tah-dah!

ref/Almost all of these results were concluded by Ramanujan, what I did is just adding more details since his paper was like, filled with "Q.E.D."


A generalization of this, you can take \( [t^q] \) in \( \sum_{n\ge 0}{\frac{(t+k)^n}{n!}\varphi_n(z)}=\sum_{n\ge 0}{\frac{t^n}{n!}\varphi_n(f^k(z))} \)

Leo

 "\( \varphi_1(z)=\lambda(z) \), which is also the difinition of the iteration velocity"

this is exactly what I am talking about with semi-group addition homomo , the conjecture

https://math.eretrandre.org/tetrationfor...p?tid=1640

https://math.eretrandre.org/tetrationfor...p?tid=1639

the escape equation and the displacement equation.

https://math.eretrandre.org/tetrationfor...p?tid=1641

In a way it was all about iteration velocity, since that imo gives the semi-group homomo.

So I will start to look at things from the julia equation perspective !!

Notice the julia equation is usually only considered for parabolic fixpoints but we can go around that.

Also the julia equation is not unique so we are not done yet.

I had the idea of connecting these ideas with andrews slog.
( and its borel transform )

Lots of work to do and mysteries to solve but im getting on a road. I have some inspiration now !


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"  A generalization of this, you can take \( [t^q] \) in \( \sum_{n\ge 0}{\frac{(t+k)^n}{n!}\varphi_n(z)}=\sum_{n\ge 0}{\frac{t^n}{n!}\varphi_n(f^k(z))} \) "

Im not sure what you mean by that ??

What is that suppose to mean ?

Is it interesting ?

Are those iteration accelerations or so ??

Im confused.


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Im thinking about generalizing Julia equation.

Julia equation is about iteration velocity and a somewhat linear displacement idea.

I was thinking about the quadratic analogue...


regards

tommy1729

In other words studying 

f( exp(x) ) = exp(x) f(x).

or maybe

f( (1 + x/n)^n ) = (1 + x/n)^n f(x)

for large n or limit n.

**

f( exp(0) ) = exp(0) f(0).

giving f(1) = f(0).

hmmm ...

f( (1+0)^n ) = f(1) = (1 + 0)^n f(0).

getting there too fast ??


I will think more b4 i post more musings.


regards

tommy1729