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 Some "Theorem" on the generalized superfunction Leo.W Junior Fellow Posts: 29 Threads: 3 Joined: Apr 2021 05/03/2021, 05:31 PM Hi everyone, I'm an amateur in iteration theory. Lately I've read one of @MphLee's posts and then my previous work occurred to me and I started to reorganize them. I've been worked for many years on what @MphLee called the generalized superfunctions (which I prefer to call "eigen-decomposit·ive functional equation"), and I have poor knowledge about the set theory so that my work may have some logistic false in it(plz someone help and I really appreciate it ). I call it "eigen-decomposit·ive" because the equation for which generalized superfunction hold true, is analogous to the eigenvalue decomposition of a matrix in the field of linear algebra, and someone may find it really analogous to conjugacy, so the name doesn't matter at all. And here I'm posting the previous part of my main work, in which I used the term "multivalued" or the concept of "multivalued function" many times, even if the term remains controversial till modern times, it(or the concept) still helps a lot in iteration theory(the reason will be explained soon, updating). And if someone have any trouble understanding it, I'd recommend to consider it as a function having many branch cuts. I hope my work could be contributive to the modern iteration theory. Regards. Attached Files Image(s)                     Leo.W Junior Fellow Posts: 29 Threads: 3 Joined: Apr 2021 05/04/2021, 02:59 AM Part 2 Attached Files Image(s)                     Leo.W Junior Fellow Posts: 29 Threads: 3 Joined: Apr 2021 05/04/2021, 03:28 AM One of sheldon's posts mentioned Peter Walker's 1991 paper https://math.eretrandre.org/tetrationfor...p?tid=1292 In which he discribed a function to connect the natural tetration and the generalized iteration of f(z)=exp(z)-1 The method is the same to Cancel Law(f(z)=z+1,g(z)=exp(z)-1,h(z)=exp(z)) Most generalized iteration for some functions having no finite fixed point, like f(z)=2sinh(ln(z)) https://math.eretrandre.org/tetrationfor...p?tid=1305 f(z)=z+z^-1 f(z)=2z+z^-1 f(z)=z+exp(z) f(z)=z+ln(z) we can use the transformation of the fixed points (or conjugacy): f(z)=2sinh(ln(z))   q(z)=z^-1   F(z)=q(f(q^-1(z)))=-1/2csch(ln(z)) having parabolic fixed point 0 f(z)=z+z^-1   q(z)=z^-1   F(z)=q(f(q^-1(z)))=z/(z^2+1) having parabolic fixed point 0 f(z)=2z+z^-1   q(z)=z^-1   F(z)=q(f(q^-1(z)))=z/(z^2+2) having elliptic(attracting) fixed point 0 f(z)=z+exp(z)   q(z)=exp(z)   F(z)=q(f(q^-1(z)))=z*exp(z) having parabolic fixed point 0 f(z)=z+ln(z)   q(z)=exp(z)   F(z)=q(f(q^-1(z)))=ln(z+exp(z)) having parabolic fixed point 0 Then we generate F^t(z) and plug it into f^t(z)=q^-1(F^t(q(z))) JmsNxn Long Time Fellow Posts: 539 Threads: 92 Joined: Dec 2010 05/05/2021, 02:59 AM Hey, Leo Welcome to the forum! It's always nice to get fresh blood. This is interesting; I'm quite the fan of the notation $\zeta(f|h) h = f\zeta(f|h)$. I'm wondering, do you have any general idea on how to define the character, $ \zeta(f|h)$ For general instances? This is something I've been stuck on; developing a general theory to handle arbitrary $f,h$; and I keep hitting dead-ends. I know MphLEE is a fan of black boxing it. I'm curious to hear what he'll have to say about this. As to conjugating fixed points, this is pretty standard, and many authors have done that before. The trouble is doing it for exotic scenarios, and not just when it's convenient. I do believe that generally finding well behaved solutions to, $ f\phi = \phi g\\$ Is an open problem. And really, the famous examples are the Schroder/Abel/Bottchner equations. Doing this, for say $\phi = \zeta(tan(z) | \exp(\exp(z)))$ would be something else though, lol. Anyway, welcome to the forum! I hope we can be of service, and we can all learn together. Regards, James Leo.W Junior Fellow Posts: 29 Threads: 3 Joined: Apr 2021 05/05/2021, 11:43 AM (This post was last modified: 05/05/2021, 06:15 PM by Leo.W. Edit Reason: Grammar correction and small changes ) (05/05/2021, 02:59 AM)JmsNxn Wrote: Hey, Leo Welcome to the forum! It's always nice to get fresh blood. This is interesting; I'm quite the fan of the notation $\zeta(f|h) h = f\zeta(f|h)$. I'm wondering, do you have any general idea on how to define the character, $ \zeta(f|h)$ For general instances? This is something I've been stuck on; developing a general theory to handle arbitrary $f,h$; and I keep hitting dead-ends. I know MphLEE is a fan of black boxing it. I'm curious to hear what he'll have to say about this. As to conjugating fixed points, this is pretty standard, and many authors have done that before. The trouble is doing it for exotic scenarios, and not just when it's convenient. I do believe that generally finding well behaved solutions to, $ f\phi = \phi g\\$ Is an open problem. And really, the famous examples are the Schroder/Abel/Bottchner equations. Doing this, for say $\phi = \zeta(tan(z) | \exp(\exp(z)))$ would be something else though, lol. Anyway, welcome to the forum! I hope we can be of service, and we can all learn together. Regards, JamesThank you, James! I'm really happy to join you all, and I'm a big fan of your respectable elaborate previous work, so impressive! The case $\phi = \zeta(tan(z) | \exp(\exp(z)))$ seems really horrible However, you can utilize the "Cancel Law" I mentioned in Section II: Letting h(z)=tan(z), f(z)=exp(exp(z)), and simply add that g(z)=z+1, then the phi{h|f} function can be represented as phi{h|g}(phi{g|f}(z)), where phi{h|g}=phi{tan(z)|z+1} is the superfunction of tan(z), and phi{g|f}=phi{z+1|exp(exp(z))} is the abel function of exp(exp(z)). The Cancel Law can produce infinitely many ways of solving the same equation by choosing different g(z). Another Law I forgot to mention is that, for function f, and some iteration f^t where t is nonzero, we can always generate one's superfunction family(abel,schroder,bottcher and so on) to another's. To see this, suppose Z satisfies Z f=g Z, then notice that Z (f^t)=(Z f) f^(t-1)=g Z f^(t-1)=g (Z f) f^(t-2)=g (g Z) f^(t-2)=g^2 Z f^(t-2)=....=(g^t) Z, so Z also satisfies the identity: Z (f^t)=(g^t) Z, unless t is 0. Then we arrive at Z=Z{g|f}~=Z{g^t|f^t}, where A~=B represents a relation that, A and B satisfies the same equation, and there's some probability that A=B (so it's also possible they're not equal). This is the ''Invariant Law'' Then we use the Cancel Law: Z{g|f^t}(z)=Z{g|g^t}(Z{g^t|f^t}(z))~=Z{g|g^t}(Z{g|f}(z)) where Z{g|g^t} is solvable, using the cancel law one more time, Z{g|g^t} can be transformed into the case g(z)=s*z then it's easy to check Z{g|g^t}=c*z^t where c in a nonzero constant. I prefer to call this the "Combination Law". I've been work on the exotic cases for so long, let's call the function f having fixed point L, and f'(L)=s. For the classic exotic case s=1, the Julia equation is always solvable in coeffients(if you assume a series of it), thus we have an approach to abel function. Then assume the asymptotic expansion of the superfunction at infinity, solve the coefficients term by term, and the superfunction is solved. And we should be careful with the direction of complex infinity, since we may generate different branch cuts of the superfunction. This method is already well-known. When I was attempting to solve the "general exotic case"/parabolic case, where s=exp(2*pi*I*q) and q is a real rational number(and not an integer), then this case has no solution able to be generated from L, which is also pretty well-known. However, let's just look on the 2-periodic case f(z)=-z(1-z), which is the logistic recurrence when s=-1: Firstly I noticed that though the fixed point 0 is unsolvable, but the function has another, f(2)=2 and f'(2)=3, hence we can generate the superfunction of f at L=2. Then I calculated that, all quadratic functions F(z)=a z^2+b z+c is always conjugate to the simplest case f(z)=z^2+v where v=a*c-b(b-2)/4, and the "conjugator" is a linear function. So I asked if all general exotic cases in f(z)=z^2+v  have another fixed point which is solvable, and it turned out this is true leaving out v=0.25. (The way to check is simply solve f(z)=z, denoting solution as z1 and z2, then use Vieta's theorem, or Newton's identities about polynomial roots, we see that f'(z1)+f'(z2)=2, so if f'(z1) is on the unit complex circle, then f'(z2) is on the outside.) Now I really wonder if there exists a function whose derivative at its fixed points are all unsolvable. The case: f(z)=z+(z-c)^n is obviously parabolic, with an nth order multi-fixed point c. Secondly I tried to calculate the asymtotic expansion of the superfunction T, generated from L=0,f(z)=-z(1-z) Applying the "Combination Law" I mentioned above, we see that, (since f(f(z))=z-2*z^3+z^4 which is classic parabolic case, we have the leading asymptotic term of the superfunction of f(f(z)), which is z^(-1/2)/2) the superfunction of f should have asymptotic expansion with leading term (2 z)^(-1/2). So I assumed that T(z)~(2 z)^(-1/2)~f((2 z)^(-1/2)), and by slight rearrangement, I used the initial guess T_0(z)=cos(pi z/2)^2/sqrt(2 z+6)+sin(pi z/2)^2*f(1/sqrt(2 z+4)) and applied T(z)=f^-1(T(z+1)) to make it converge. Though the function is rather converging to 0 than converging to a smooth function, I plot the T function after 50 iterations, T_50(z)=f^-51(f(T_0(z+50))), and the function did satisfy the relation: T(z+1)=f(T(z)) up to about 7 decimal places. So I guess the superfunction can be generated but in a rather hard method? Lastly here's the plot of the T_50 function The image is exported by the Wolfram Mathematica function AbsArgPlot. (I'm sorry for my poor English and if I did say some words that is offensive or impolite, I'm really sorry for that) (ps I wonder how you type a math formula in your reply? could u plz help me in it?) Regards Leo Attached Files Image(s)     JmsNxn Long Time Fellow Posts: 539 Threads: 92 Joined: Dec 2010 05/05/2021, 07:53 PM Hey, Leo, very interesting stuff. Don't worry about your English, you speak very well; it's alright. To add in Latex on this forum just write Code:$$math goes here$$ As to what you do with $\zeta(\tan(z)|\exp(\exp(z)))$; that's very clever. That's particular to what MphLee and I were talking about. I guess my question was more so, how do you plan to choose which superfunction we produce? That was more what I was asking.  I get that, $ \tan(F(z)) = F(z+1)\\ \exp(\exp(G(z))) = G(z+1)\\ \zeta(\tan(z)|\exp(\exp(z))) = F(G^{-1}(z))\\ $ I was wondering if you had any method where it chooses a particular super function $F$ and $G$. Regards, James Leo.W Junior Fellow Posts: 29 Threads: 3 Joined: Apr 2021 05/06/2021, 05:37 AM (05/05/2021, 07:53 PM)JmsNxn Wrote: Hey, Leo, very interesting stuff. Don't worry about your English, you speak very well; it's alright. To add in Latex on this forum just write Code:$$math goes here$$ As to what you do with $\zeta(\tan(z)|\exp(\exp(z)))$; that's very clever. That's particular to what MphLee and I were talking about. I guess my question was more so, how do you plan to choose which superfunction we produce? That was more what I was asking.  I get that, $ \tan(F(z)) = F(z+1)\\ \exp(\exp(G(z))) = G(z+1)\\ \zeta(\tan(z)|\exp(\exp(z))) = F(G^{-1}(z))\\ $ I was wondering if you had any method where it chooses a particular super function $F$ and $G$. Regards, James Thanks a lot, James!  (Maybe here's about how I choose a superfunction, I don't know if I expressed it in a clear way?) I suggest we denote $\alpha\{f(z)\}(z)$ as the specific abel function of f(z), and $\sigma\{f(z)\}(z)$ as the schroder function and so on. Straightforwardly, the superfunction $G$ is exactly (probably) equal to $\alpha^{-1}\{\exp(z)\}(\frac{z}{2})$. Choosing the appropriate branch cut of  $\alpha^{-1}\{tan(z)\}(z)$ is kind of tricky. First of all, we should choose a branch cut which has the largest range, so an entire cut that contains a whole complex plane is better than a cut is either not entire or its range is only a subset of $\mathbb{C}$, the advantage of this method is that the $\zeta\{\tan(z)|\exp(\exp(z))\}$ we want would be entire, otherwise it may contain branch cuts or undefined for some values. For instance, the superfunction of sine function has infinitely many branch cuts, the most common or most times generated one is that $T(z)=\alpha^{-1}\{sin(z)\}(z)=\sqrt{\frac{3}{z}}+\mathrm{o}(\frac{1}{z})$. By iterating $T(z)=\sin^{-1}(T(z+1))$ we have this branch cut. However it is not entire, in fact the range of this function contains no negetive real numbers. The range is about to stay in the area approximately $\arg(z)\in[-\frac{\pi}{3},\frac{\pi}{3}]$. Another branch cut which is pretty approachable is with the initial condition $T(0)=i$, which is connected to the superfunction of sinh(z): $\alpha^{-1}\{sin(z)\}(z)=i\alpha^{-1}\{\sinh(z)\}(z)$, and if this cut is an entire function, we'll choose it instead of the former one. However, for some reason we have to choose those non-entire cuts, if you want to generate a real-to-real $\zeta$, sometimes it's inevitable to use such functions. For a real-valued function, it's better to choose a real-to-real superfunction than a real-to-complex superfunction, e.g. the$G^{-1}(z)=\frac{1}{2}\alpha\{\exp(z)\}(z)$, where we use the merged version of $\alpha\{\exp(z)\}(z)=slog_e(z)$. So the order of what kind of superfunctions I choose is, if f is real-to-real, T: real-to-real(entire,merged)>real-to-complex(entire,merged)>real-to-complex(entire,unmerged) >real-to-real(not entire)>real-to-complex(not entire) if f is real-complex, T:real-to-complex(entire, range is the whole plane,merged)>real-to-complex(entire, range is subset of C)>real-to-complex(not entire) And for a function, its superfunction predicted to have an oscillating behavior(such as 0.1^x), real-to-complex(entire,merged) is the best choice to use. For the merged function, we use the very 2 fixed points having the least absolute value and larger real part. For real-to-real function, we'd better use the fixed point whose multipler s>1, since 0

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