Inverse super-composition - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Inverse super-composition (/showthread.php?tid=1104) Pages: 1 2 Inverse super-composition - Xorter - 11/24/2016 Let us invastigate the composition and its iterated function. We know the followings: (f o g) o g^-1 = f f^-1 o (f o g) = g ☉ f ☥^N = f o f o ... o f (N times), where ☉ is called steinix and ☥ is called ankh. ☉ f ☥^N o ☉ f ☥^M = ☉ f ☥^(N+M) ☉(☉ f ☥^N)☥^M = ☉ f ☥^(N*M) f o ☉ f ☥^N = ☉ f ☥^N o f = ☉ f ☥^(N+1) ... etc. The question is that what the inverses of the ☉ f(x) ☥^N, steinix-ankh formula is? According to the previous rules, I can find one of the inverses which is the next: ☉(☉ f ☥^N)☥^(1÷N) = f But I am interested in that what the other inverse is which would give me N. So ☉ f ☥^N {something operator}(f) = N, what is it? (It might be called steinix-logarithm.) Any thoughts? RE: Inverse super-composition - JmsNxn - 11/25/2016 This is just the abel function which satisfies $\alpha(f(x)) = \alpha(x) + 1$ and then if $\alpha_x(z) = \alpha(z) - \alpha(x)$ we have $\alpha_x(f^{\circ n}(x)) = n$ RE: Inverse super-composition - Xorter - 12/23/2016 Hmm. And what is the taylor series of it? Or what else could we know about this function? RE: Inverse super-composition - JmsNxn - 12/23/2016 (12/23/2016, 01:33 PM)Xorter Wrote: Hmm. And what is the taylor series of it? Or what else could we know about this function? Taylor series is a mess, can usually only be solved in specific instances. There are more natural ways of representing it other than Taylor series. Ecalle has a nice formula, but I can't remember it. RE: Inverse super-composition - Xorter - 12/24/2016 (12/23/2016, 08:12 PM)JmsNxn Wrote: (12/23/2016, 01:33 PM)Xorter Wrote: Hmm. And what is the taylor series of it? Or what else could we know about this function? Taylor series is a mess, can usually only be solved in specific instances. There are more natural ways of representing it other than Taylor series. Ecalle has a nice formula, but I can't remember it. I am interested in all the working methods, it can be easy way or not. Please, if you do not mind, it would be really helpful for me and for the community if you tried to find it. Thank you very much. RE: Inverse super-composition - sheldonison - 12/25/2016 (12/24/2016, 09:53 PM)Xorter Wrote: I am interested in all the working methods, it can be easy way or not. Please, if you do not mind, it would be really helpful for me and for the community if you tried to find it. Thank you very much. Your question is too general, since you don't identify what f(x) you are interested in. In general, the type of solution depends on the behavior at the fixed point. I assume you are interested in real valued functions. Some iterated functions have an attracting point. Then we look at the slope at the fixed point. If the slope at the fixed point is equal to 1, then we have the parbolic case, which is where the method of Ecalle works. This is the method that James was refering too. Ecalle's method can be used to find the solution for the slog inverse of the iterated function for $f(z) \mapsto \eta^z$ where $\eta=\exp(1/e)$, and the fixed point is "e". Then the method of Ecalle generates the Abel function at the fixed point. See http://mathoverflow.net/questions/45608/does-the-formal-power-series-solution-to-ffx-sin-x-converge and look for the $\alpha(z)$ formal power series definition where Will Jagy writes, "Now, given a specific x....it is a result of Jean Ecalle at Orsay that we may take". The algebra for the method of Ecalle is easiest if the fixed point is moved to zero, by solving the Abel function for the equivalent probem, $f(y) \mapsto \exp(y)-1$ instead of $f(z) \mapsto \eta^z$ where $y=\frac{z}{e}-1$ If the slope is less than 1, then we can use Koenig's Schröder's function solution; see https://en.m.wikipedia.org/wiki/Schröder's_equation If there are no real valued fixed points, then we have Kneser's solution for tetration. There are various numerical solutions for Kneser's slog such as mine: http://math.eretrandre.org/tetrationforum/showthread.php?tid=1017 RE: Inverse super-composition - Xorter - 12/25/2016 (12/25/2016, 04:16 AM)sheldonison Wrote: Your question is too general, since you don't identify what f(x) you are interested in. In general, the type of solution depends on the behavior at the fixed point. I assume you are interested in real valued functions. Some iterated functions have an attracting point. Then we look at the slope at the fixed point. Yes, my question is general, because I am looking for a totally general solution for all the kind of problem like this. Anyway, I found a nicer formula for my question instead of steinix-ankh, like this: $f(x) ^ o ^N = g(x)$ According to the knowledge of f and g, what is N? How can I calculate it? For example: $2x ^ o ^N = x^2$ Thus N must be = log2(x), but here is the question why and how can I know it from? RE: Inverse super-composition - sheldonison - 12/25/2016 (12/25/2016, 04:38 PM)Xorter Wrote: (12/25/2016, 04:16 AM)sheldonison Wrote: Your question is too general, since you don't identify what f(x) you are interested in. In general, the type of solution depends on the behavior at the fixed point. I assume you are interested in real valued functions. Some iterated functions have an attracting point. Then we look at the slope at the fixed point. Yes, my question is general, because I am looking for a totally general solution for all the kind of problem like this. Anyway, I found a nicer formula for my question instead of steinix-ankh, like this: $f(x) ^ o ^N = g(x)$ According to the knowledge of f and g, what is N? How can I calculate it? For example: $2x ^ o ^N = x^2$ Thus N must be = log2(x), but here is the question why and how can I know it from? For a fixed point of zero, with a fixed point multiplier of 2, the general solution for the Abel function generated at the fixed point of zero is: $f(z) = 2x + \sum_{n=2}^{\infty}a_n z^n$ $\alpha(z) = \log_2(S(z))\;\;\;$ This is the Abel function for f(z) $\;\alpha(f(z)) = \alpha(z)+1$ where S(z) is the formal Schröder equation solution; $S(f(z)) = 2\cdot S(z)\;\;\; S(z)=z+\sum_{n=2}^{\infty}b_n z^n\;\;$ This is sometimes called Koenig's solution. It can be modified to work with any fixed point multiplier of k, |k|<>1. Using pari-gp one can easily write a program to generate the formal power series for S(x) given f(x). RE: Inverse super-composition - Xorter - 12/25/2016 (12/25/2016, 08:35 PM)sheldonison Wrote: For a fixed point of zero, with a fixed point multiplier of 2, the general solution for the Abel function generated at the fixed point of zero is: $f(z) = 2x + \sum_{n=2}^{\infty}a_n z^n$ $\alpha(z) = \log_2(S(z))\;\;\;$ This is the Abel function for f(z) $\;\alpha(f(z)) = \alpha(z)+1$ where S(z) is the formal Schröder equation solution; $S(f(z)) = 2\cdot S(z)\;\;\; S(z)=z+\sum_{n=2}^{\infty}b_n z^n\;\;$ This is sometimes called Koenig's solution. It can be modified to work with any fixed point multiplier of k, |k|<>1. Using pari-gp one can easily write a program to generate the formal power series for S(x) given f(x). I almost undestand it. Okey, than what is the a[n] and b[n] in the sum formula? I feel we are closer then ever before. Could you show me this way with another example, please? For instance, let us invastigate it: $cos ^o ^N (x) = sin(x)$. (And cos' fixed point is ~0.739.) The question is that what N is and how I can calculate it. RE: Inverse super-composition - sheldonison - 12/26/2016 (12/25/2016, 10:23 PM)Xorter Wrote: (12/25/2016, 08:35 PM)sheldonison Wrote: For a fixed point of zero, with a fixed point multiplier of 2, the general solution for the Abel function generated at the fixed point of zero is: $f(z) = 2x + \sum_{n=2}^{\infty}a_n z^n$ $\alpha(z) = \log_2(S(z))\;\;\;$ This is the Abel function for f(z) $\;\alpha(f(z)) = \alpha(z)+1$ where S(z) is the formal Schröder equation solution; $S(f(z)) = 2\cdot S(z)\;\;\; S(z)=z+\sum_{n=2}^{\infty}b_n z^n\;\;$ This is sometimes called Koenig's solution. It can be modified to work with any fixed point multiplier of k, |k|<>1. Using pari-gp one can easily write a program to generate the formal power series for S(x) given f(x). I almost undestand it. Okey, than what is the a[n] and b[n] in the sum formula? I feel we are closer then ever before. Could you show me this way with another example, please? For instance, let us invastigate it: $cos ^o ^N (x) = sin(x)$. (And cos' fixed point is ~0.739.) The question is that what N is and how I can calculate it. $L \approx 0.73908513322;$ $f(x) = \cos(x+L)-L = \cos(L)\cos(x)-\sin(L)\sin(x)-L$ $f(x) \approx -0.67361202918x-0.36954256661x^2+0.11226867153x^3...$ I make no attempt to address your question, partly because I doubt your question has an answer. Using f(x) as an example to explore the Schröder equation probably isn't a good first example either because the multiplier at the fixed point, which is the a1 coefficient of f(x), is negative. Therefore it is no longer possible for the fractional iterates of the function $f^{[\circ x]$ to be real valued. So the Taylor series of the Schröder equation for f(x) isn't relevant either, but it follows the definition from my previous post. Here are the first couple of terms. ${S(x)\approx x+ 0.32779313060x^2-0.74862437767x^3...$