08/10/2022, 12:47 PM
I. Distincts among members in generalized superfunction families
We present a generalized superfunction family by MphLee's notation \([f,g]\) or my previously used notation \(\zeta{g|f}(z)\), and by previous discussion we have \([g,g][f,g]=[f,g][f,f]=[f,g]\), for a specific \(f(z)=z+1\), we denote \(\theta=[f,f]\) as \(\theta(z+1)=\theta(z)+1\), and can be written in the form \(\theta(z)=z+c+T(z)\) where T(z) is a 1-periodic function of z.
It's natural to consider to distinct each member, to figure out on which theta mapping the function depends is the key. But also we should consider that, the theta mapping may not map among all members, well in theoretical frames it does, but in pragmatic computations no, it's easy to check for example that 2 schroder superfunctions (explicitly discussed by James and MphLee in a previous post) generated at ~0.3+1.3i and ~0.3-1.3i won't be able to transform each into the other by one single computable theta-mapping. So we can consider the superfunction sub-families for different branch cuts at different fixed points. It's pretty easy to distinct sub-families from sub-families.
And then under each sub-family, the way to distinct each member is theta mapping. It's pretty hard though. However, because we almost never use of a fixed point which has a multiplier 1, after theta mapping, the superfunction would have (at least looks so) 2 different periods (and not elliptic functions), or 1 as a fake period. Then we can detect a function by some sort-of Fourier transformation to check whether it has a theta mapping or not, and distinct then.
As we wrote, \(F(z+1)=f(F(z)), F(z)=\sigma^{-1}(L( c)s^z)\) where c refers to that in the \(\theta(z)=z+c+T(z)\), and a theta mapping-ed F is \(F_\theta(z)=\sigma^{-1}(L_\theta( c)s^{z+T(z)})\) and then we can write such "Fourier transformation" to take T out. Even the case we don't know about the original sigma function we can still detect the 1-periodic behavior.
II. A special functional equation and related phenomenon(?)
I personally got into some sorts of asymptotic solutions of iterative functions, one of which is \(\Delta{f}=g\circ{f}\) for given function g. This is exactly asking the superfunction \(f(z+1)=G(f(z))\) where \(G(z)=z+g(z)\), However I don't need an exact solution but just the first term of the asymptotic expansion, and by the superfunction method I can get an exact solution by iteration like \(f_n(z)=G^{-n}(f_0(z+n))\).
And easily to check if g(z)=O(z), only the biggest term of g(z) controls the biggest term of asymp of f. For example, if g(z)=1/z, we can write for \(G(z)=z+\frac{1}{z}\), and then both \(G^t(z)=z+\frac{t}{z}+O(t^2z^{-2})\) and \(G^t(z)=\sqrt{2t}+O(t^0h(z))\) for some h(z) are true.
The sqrt term comes directly from the equation.
Another example, for \(G(z)=z+\sqrt{z}\) we have \(G^t(z)=\frac{t^2}{4}+O(h(z)t\log(t)^2)\)
I tried a^(-z), z^a*log(z)^b(a<1), etc.
For g(z)=O(z), can we always derive such terms for a specific asymptotic term O(g(z))?
We present a generalized superfunction family by MphLee's notation \([f,g]\) or my previously used notation \(\zeta{g|f}(z)\), and by previous discussion we have \([g,g][f,g]=[f,g][f,f]=[f,g]\), for a specific \(f(z)=z+1\), we denote \(\theta=[f,f]\) as \(\theta(z+1)=\theta(z)+1\), and can be written in the form \(\theta(z)=z+c+T(z)\) where T(z) is a 1-periodic function of z.
It's natural to consider to distinct each member, to figure out on which theta mapping the function depends is the key. But also we should consider that, the theta mapping may not map among all members, well in theoretical frames it does, but in pragmatic computations no, it's easy to check for example that 2 schroder superfunctions (explicitly discussed by James and MphLee in a previous post) generated at ~0.3+1.3i and ~0.3-1.3i won't be able to transform each into the other by one single computable theta-mapping. So we can consider the superfunction sub-families for different branch cuts at different fixed points. It's pretty easy to distinct sub-families from sub-families.
And then under each sub-family, the way to distinct each member is theta mapping. It's pretty hard though. However, because we almost never use of a fixed point which has a multiplier 1, after theta mapping, the superfunction would have (at least looks so) 2 different periods (and not elliptic functions), or 1 as a fake period. Then we can detect a function by some sort-of Fourier transformation to check whether it has a theta mapping or not, and distinct then.
As we wrote, \(F(z+1)=f(F(z)), F(z)=\sigma^{-1}(L( c)s^z)\) where c refers to that in the \(\theta(z)=z+c+T(z)\), and a theta mapping-ed F is \(F_\theta(z)=\sigma^{-1}(L_\theta( c)s^{z+T(z)})\) and then we can write such "Fourier transformation" to take T out. Even the case we don't know about the original sigma function we can still detect the 1-periodic behavior.
II. A special functional equation and related phenomenon(?)
I personally got into some sorts of asymptotic solutions of iterative functions, one of which is \(\Delta{f}=g\circ{f}\) for given function g. This is exactly asking the superfunction \(f(z+1)=G(f(z))\) where \(G(z)=z+g(z)\), However I don't need an exact solution but just the first term of the asymptotic expansion, and by the superfunction method I can get an exact solution by iteration like \(f_n(z)=G^{-n}(f_0(z+n))\).
And easily to check if g(z)=O(z), only the biggest term of g(z) controls the biggest term of asymp of f. For example, if g(z)=1/z, we can write for \(G(z)=z+\frac{1}{z}\), and then both \(G^t(z)=z+\frac{t}{z}+O(t^2z^{-2})\) and \(G^t(z)=\sqrt{2t}+O(t^0h(z))\) for some h(z) are true.
The sqrt term comes directly from the equation.
Another example, for \(G(z)=z+\sqrt{z}\) we have \(G^t(z)=\frac{t^2}{4}+O(h(z)t\log(t)^2)\)
I tried a^(-z), z^a*log(z)^b(a<1), etc.
For g(z)=O(z), can we always derive such terms for a specific asymptotic term O(g(z))?
Regards, Leo 