Some "Theorem" on the generalized superfunction

[Leo.W]

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

Thank you James

A quick proof is: (no need for complex algebra)

Assume the iteration can be expanded as
\( f^t(z)=\sum_{n\ge 0}{\frac{t^n}{n!}\varphi_n(z)} \) (1)
First note that \( \frac{\mathrm{d}f^k(z)}{\mathrm{d}k}=\sum_{n\ge 0}{\frac{nk^{n-1}}{n!}\varphi_n(z)} \) (2)
We then take the fundamental definition \( f^{t+k}(z)=f^t(f^k(z)) \)
Expand the LHS by (1) and consider \( f^k(z) \) be the variable of RHS, then expand by (1)
We have now: \( \sum_{n\ge 0}{\frac{(t+k)^n}{n!}\varphi_n(z)}=\sum_{n\ge 0}{\frac{t^n}{n!}\varphi_n(f^k(z))} \)
Take the coefficient of \( [t^1] \), LHS by binomial theorem, and compare with (2)
we have \( \frac{\mathrm{d}f^k(z)}{\mathrm{d}k}=\varphi_1(f^k(z)) \) (3)
Set \( y=f^k(z),x=f^k(y) \)
Use chain rule to show \( \frac{\mathrm{d}x}{\mathrm{d}k}=\frac{\mathrm{d}x}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}k} \)
Or plug in, so we have \( \frac{\mathrm{d}f^k(y)}{\mathrm{d}k}=\frac{\mathrm{d}f^k(y)}{\mathrm{d}y}\frac{\mathrm{d}f^k(z)}{\mathrm{d}k} \)
by (3), we can now show that \( \frac{\mathrm{d}f^k(y)}{\mathrm{d}k}=\frac{\mathrm{d}f^k(y)}{\mathrm{d}y}\varphi_1(y) \) (4)
(now substitute y with z)This shows we can calculate all \( \varphi_n(z) \) by equating both's coefficients of \( [t^n] \)
followed by \( \varphi_n(z)=\varphi_1(z)\frac{\mathrm{d}\varphi_{n-1}(z)}{\mathrm{d}z} \)
Then we only need to determine \( \varphi_1(z) \), then we can construct the iteration in a neighborhood of t=0, which can be considered as the infinitesimal iteration

We also need the basic fact that Julia's solution works for all iterations of the same function, as stated:
Denote \( \lambda(f(z))=\lambda(z)\frac{\mathrm{d}f(z)}{\mathrm{d}z} \) with Julia's solution \( \lambda(z) \)
Then \( \lambda(f^k(z))=\lambda(z)\frac{\mathrm{d}f^k(z)}{\mathrm{d}z} \) (5)
which can be proved very easily by law 3, or through abel's solution

Now note the integral (it's exactly the Abel's solution though: \( \alpha'(z)\lambda(z)=1 \)) q is arbitrary, and z is not a fixed point, but can be very close to some fixed point L

\( \int_{f^{q-1}(z)}^{f^{q}(z)}\frac{dz}{\lambda(z)}=\int_{f^{q-1}(z)}^{f^{q}(z)}\frac{dz}{(\frac{\lambda(f(z))}{f'(z)})}=\int_{f^{q-1}(z)}^{f^{q}(z)}\frac{f'(z)dz}{\lambda(f(z))}=\int_{f^{q-1}(z)}^{f^{q}(z)}\frac{d(f(z))}{\lambda(f(z))}=\int_{f^{q}(z)}^{f^{q+1}(z)}\frac{dz}{\lambda(z)}=C \)

so we have now the more general integral, showing that \( \int_{z}^{f^{k}(z)}\frac{dz}{\lambda(z)}=kC \) (6)

We should consider the integral here as a contour integral, so that (wherever lambda is holomorphic, we can take it as a line integral), if somewhere lambda had a branch cut, we can then know that this equation defines a RS.
By limit, we can analytically continue the definition, so then z can be some fixed point L, however, we should then re-consider f as a multivalued function so that by Cauchy's residue theorem, there lies some branch cuts near L and L is a singular, so C is then not 0.
Now take derivative of (6) with respect to k:

\( \frac{\mathrm{d}}{\mathrm{d}k}\int_{z}^{f^{k}(z)}\frac{dz}{\lambda(z)}=\frac{\mathrm{d}}{\mathrm{d}k}(\alpha(f^k(z))-\alpha (z))=\frac{\mathrm{d}(\alpha(f^k(z)))}{\mathrm{d}k}=\frac{\mathrm{d}(\alpha(f^k(z)))}{\mathrm{d}f^k(z)}\frac{\mathrm{d}f^k(z)}{\mathrm{d}k}(LHS)=C(RHS) \)

Combine this with (4) and (5) we have \( \frac{\mathrm{d}(\alpha(f^k(z)))}{\mathrm{d}f^k(z)}\frac{\mathrm{d}f^k(z)}{\mathrm{d}k}=\frac{1}{\lambda(f^k(z))}\frac{\mathrm{d}f^k(z)}{\mathrm{d}z}=\frac{1}{\lambda(z)\frac{\mathrm{d}f^k(z)}{\mathrm{d}z}}\frac{\mathrm{d}f^k(z)}{\mathrm{d}k}(LHS)=C(RHS) \)
In conclusion, \( \frac{1}{\lambda(z)\frac{\mathrm{d}f^k(z)}{\mathrm{d}z}}\frac{\mathrm{d}f^k(z)}{\mathrm{d}k}=C \) (7)
Try to remind of the law 3, saying that by multiplying a constant, one can modulate absolute value of the Abel's solution, then we can set C=1

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