The iterational paradise of fractional linear functions bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 08/05/2022, 12:34 AM (This post was last modified: 08/05/2022, 12:54 AM by bo198214. Edit Reason: wording ) And now to the culmination of two-fixed points regular iteration: parabolic plosion! We have quite a similar situation as with the exponentials where the base transits from $$b<\eta$$ to $$b>\eta$$. But this time with the fractional linear function: $$f(z) = \frac{c+z}{1+z}$$ The fixed points are $$\pm\sqrt{c}$$. By varying c from positive to negative the two fixed points on the real axis unite at $$c=0$$ being a parabolic fixed point and then turn into conjugate complex fixed points. Everything is very similar here to moving $$b$$ from $$<\eta$$ to $$>\eta$$ for exponentials $$b^z$$, EXCEPT that this is the perfect world! Everything we had wished for the exponentials like: * iteration at the lower fixed point is the same function as iteration at the upper fixed point * iterating at one of the complex fixed points would give real values on the real axis * crescent iteration is regular iteration * iteration at the parabolic fixed point is analytic there * $$b\mapsto \exp_b^{\circ t}(z)$$ is analytic at $$\eta$$ is true here! (Maybe a bit too perfect world like the Witnesses of Jahwe would imagine!) So we again have an explicit formula for the iteration of $$f$$, however it is a bit more difficult too derive and I will leave it to you to check this formula (for $$f^{\circ s+t} = f^{\circ s}\circ f^{\circ t}$$ or even derive it yourself.) $$f^{\circ t}(z) = \frac{q_t(\sqrt{c})c + z}{1+q_t(\sqrt{c})z}, q_t(s)=\frac{(1+s)^t-(1-s)^t}{s((1+s)^t+(1-s)^t)}$$ interesting to find out that $$q_t(\sqrt{c})$$ is always real for real $$c,t$$, particularly for $$c<0$$ where $$s=\sqrt{c}$$ is imaginary. This comes from the series expansion $$(1+z)^t = \sum_{n=0}^\infty \binom{t}{n} z^n$$ Then \begin{align} (1+z)^t - (1-z)^t &= 2\sum_{n=0}^\infty \binom{t}{2n+1} z^{2n+1}\\ (1+z)^t + (1-z)^t &= 2\sum_{n=0}^\infty \binom{t}{2n} z^{2n}\\ (1+ix)^t - (1-ix)^t &= 2i\sum_{n=0}^\infty \binom{t}{2n+1} (-1)^n x^{2n+1}\\ (1+ix)^t + (1-ix)^t &= 2\sum_{n=0}^\infty \binom{t}{2n} (-1)^n x^{2n} \end{align} All in all $$q_t(s)$$ is that's why real valued for real and imaginary $$s$$. Also something that is not immediately visible: $$t\mapsto q_t(s)$$ is a periodic function for imaginary $$s$$. I have to admit that I don't see how to prove this directly, anyone any ideas? Here a picture of $$t\mapsto q_t(\sqrt{c})$$  for $$c=-0.5$$. It reminds me quite of tan:     On the other hand this is something we know already from regular iteration: it is periodic in t with period $$\frac{2\pi i}{\log(d)}$$ where d is the derivative at the fixed point it is developed on. We counter check: \begin{align} f'(z) &= -\frac{c + z}{(z + 1)^2} + \frac{1}{z + 1} \\ f'(\sqrt{c}) &= -\frac{c+\sqrt{c}}{(\sqrt{c}+1)^2} + \frac{1}{\sqrt{c}+1} = \frac{1-c}{(1+\sqrt{c})^2} \end{align} That $$\frac{2\pi i}{\log(f'(\sqrt{c}))}$$ is even real is already difficult to see, we need the power series development of log: $$\log(1+z) = \sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n}$$ Then we use this with our above expression: $$\log\left(\frac{1-c}{(1+\sqrt{c})^2}\right) = \log(1-c) - 2\log(1+\sqrt{c})$$ We show that the expression on the right side is purely imaginary. To make it clearer that we are talking about $$c<0$$, I substitute $$c = (ix)^2$$: \begin{align}   \log(1+x^2) - 2\log(1+ix) &= \sum_{n=1}^\infty (-1)^{n-1}\frac{ x^{2n}}{n} - 2\sum_{n=1}^\infty (-1)^{2n-1} \frac{(ix)^{2n}}{2n} - 2 \sum_{n=0}^\infty (-1)^{2n} \frac{(ix)^{2n+1}}{2n+1}\\ &= \sum_{n=1}^\infty (-1)^{n-1}\frac{ x^{2n}}{n} - \sum_{n=1}^\infty (-1) (-1)^n\frac{(x)^{2n}}{n} - 2i \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}\\ &= - 2i \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} \end{align} How can such a simple statement about an expression being real/imaginary, be so tedious to show! Ok, but from this follows that the period $$\frac{2\pi i}{\log(1-c) - 2\log(1+\sqrt{c})}$$ is real for $$c<0$$ and that it is the same period (though negative) at the other fixed point $$-\sqrt{c}$$. Also interesting to check the limit for $$s=0$$! This is the parabolic case from my previous posts: $$\lim_{s\to 0} q_t(s) = t$$ because \begin{align} q_t(z)=\frac{2\sum_{n=0}^\infty \binom{t}{2n+1} z^{2n+1}}{2z\sum_{n=0}^\infty \binom{t}{2n} z^{2n}} = \frac{z(\binom{t}{1}  + z r_1(t))}{z(1+z r_2(t))} = \frac{t  + z r_1(t)}{1+z r_2(t)} \end{align} From this representation even follows that $$q_t(s)$$ is analytic at $$s=0$$! And so is our original function: $$f_s^{\circ t}(z) = \frac{q_t(\sqrt{c})c + z}{1+q_t(\sqrt{c})z} = \frac{q_t(s)s^2 + z}{1+q_t(s)z}$$ analytic in $$s$$ at $$0$$. And here the perfect world from regular iteration of linear fractional functions as motion picture:     bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 08/05/2022, 12:42 AM The bon mot for JmsNxn is that for $$c<0$$ at both conjugated fixed points there is a vicinity independent of $$t$$ where the iterates are holomorphic (didnt you call that "local iteration"?), because the pole is always on the real axis. But no fear James! There are other essential drawbacks that this solution has  I will post a picture later ... JmsNxn Ultimate Fellow Posts: 985 Threads: 117 Joined: Dec 2010 08/05/2022, 01:20 AM Oooooooooo very excited by this. I think I see how you can derive that the function $$q_t(s)$$ is periodic in $$s$$ too (I'm assuming for fixed $$t$$, but the period depends on $$t$$). I wrote a bunch but I scrapped it because I had assumed something that was false, but it's looking like it should be inverse mellin transformable/differintegrable. And then you can use that since $$q_n(s)$$ is constant (hence periodic (or else you wouldn't say that)--no rational function is periodic unless it's constant) to derive periodicity for each $$t$$, but I'm not certain yet. Also, I think this would only qualify as a local iteration if we allowed poles. Where then we are talking about functions on $$\widehat{\mathbb{C}}$$. So a local iteration about two points independent of $$t$$ will probably be possible, but once extended to its maximal domain will have poles. There's a reason $$\widehat{\mathbb{C}}$$ is treated as its own beast in Milnor (he calls it spherical), it changes the game entirely. The euclidean case, is specific to transcendental entire functions; that have rules of their own. So I'm still not convinced that, in the euclidean case, you've provided a counter example. But still, you are widening my eyes as to what I've assumed ! Also, I don't think you even need to put in as much work as you've done to find a counter example. Taking conjugations of $$\lambda^t z$$ by linear fractional transformations, will probably already supply that. I'll need to work harder to carve out exactly what the theorem to be. Namely, that it should be only specific to Euclidean transformations (Transcendental entire functions). bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 08/05/2022, 11:22 AM (This post was last modified: 08/05/2022, 11:51 AM by bo198214.) (08/05/2022, 01:20 AM)JmsNxn Wrote: Taking conjugations of $$\lambda^t z$$ by linear fractional transformations, will probably already supply that. This really raises a point that always puzzled me: In the calculations of the linear fractional iterations there naturally always occur the powers of *two* Eigenvalues $$\lambda_1^t$$ and $$\lambda_2^t$$. I was wondering how they relate to the *one* $$\lambda$$ you are talking about and how they relate to the two fixed points and their derivations. So I write it down to make it clear for me: So we have a linear fractional function $$\phi$$ and its matrix representation $$M$$. $$M$$ can be diagonalized in two ways with $$\lambda_1,\lambda_2$$ being the Eigenvalues of the characteristic polynomial: \begin{align} M &= A_1 D_1 A_1^{-1}, \quad\quad D_1=\left(\begin{matrix}\lambda_1 &0\\0 &\lambda_2\end{matrix}\right)\\ M &= A_2 D_2 A_2^{-1}, \quad\quad D_2=\left(\begin{matrix}\lambda_2 &0\\0 &\lambda_1\end{matrix}\right) \end{align} And because these are only permutations in columns and rows we know that both iterations are the same $$M^t = A_1 D_1^t A_1^{-1} = A_2 D_2^t A_2^{-1}$$ $$A_i$$ and $$D_i$$ correspond to the linear fractional functions $$\mu_i$$ and $$\ell_i$$ $$\phi = \mu_i \circ \ell_i \circ \mu_i^{-1}, \quad\quad\ell_i(z) = c_i z$$ $$\mu_i$$ are (obviously?) the mappings which map $$\{0,\infty\}$$ to the fixed points $$\{z_1,z_2\}$$. Say $$\mu_1(0)=z_1$$, $$\mu_2(0)=z_2$$. From this viewpoint we also can derive equality: knowing that there can only be one Schröder iteration at a fixed point and the iterations are analytic at both fixed points and that's why the iterations are equal. And it turns now out how the fixed point derivations are connected with the Eigenvalues:  $$c_1 = \frac{\lambda_1}{\lambda_2}$$ and $$c_2 = \frac{\lambda_2}{\lambda_1}$$ EDIT: and it makes totally sense that it is a ratio, because the matrix $$M$$ is only determined up to a constant a. If $$M$$ corresponds to $$\phi$$ then also $$aM$$ corresponds to $$\phi$$. Though we still have to show that $$c_i$$ are really the fixed point derivations of $$\phi$$: \begin{align} \phi' &= \mu'\circ \ell\circ\mu^{-1} \cdot \ell'\circ \mu^{-1} \cdot \frac{1}{\mu'\circ \mu^{-1}}\\ \phi'(z_i) &= \mu_i'(\ell_i(0)) \cdot \ell_i'(0) \cdot \frac{1}{\mu_i'(0)} = \ell_i'(0) = c_i \end{align} So the $$c_1=\phi'(z_1)$$ and $$c_2=\phi'(z_2)$$ are indeed the derivatives at both fixed points and they are reciprocal. So in my constructed parabolic plosion one could express $$q_t(s)$$ also in terms of one fixed point derivative $$d=\frac{1-s}{1+s}$$: $$q_t(s) = \frac{(1+s)^t - (1-s)^t}{s((1+s)^t+(1-s)^t} = \frac{1-d^t}{s\left(1+d^t\right)}$$ And in this form it is quite natural to show periodicity: Because $$d^t$$ has obviously period $$2\pi i/\log(d)$$ and we showed already that $$\log(d)$$ is purely imaginary for s imaginary. Even that is natural to show with this simpler form of the derivative d (which I didn't see in my previous post). bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 08/06/2022, 08:38 AM (This post was last modified: 08/06/2022, 08:43 AM by bo198214.) OMG I just found out it *is* based on the $$\tan$$. These are well known identities: $$\tanh(x)=\frac{e^{2x}-1}{e^{2x}+1},\quad \text{artanh}(x)={\frac {1}{2}}\log \left({\frac {1+x}{1-x}}\right),\quad i\tan(x) = \tanh(ix)$$ I am digging out long forgotten high school math, lol. In our (new) formula  ( remember $$d=\frac{1-s}{1+s}$$ ): $$q_t(s) = \frac{1}{s}\frac{1-d^t}{1+d^t} = -\frac{1}{s}\frac{e^{t\log(d)}-1}{e^{t\log(d)}+1} = -\frac{1}{s}\tanh\left(t\frac{\log(d)}{2}\right) = -\frac{1}{s}\tanh\left(-t\;\text{artanh}(s)\right) = \frac{1}{s}\tanh(t\;\text{artanh}(s))$$ And if now s is imaginary, $$s=ix$$, then $$q_t(ix) = \frac{1}{ix}\tanh\left(t\;\text{artanh}(ix)\right) = \frac{1}{x}\tan\left(\frac{t}{i}i\arctan(x)\right) = \frac{1}{x}\tan\left(t\;\arctan(x)\right)$$ And from here its totally clear that $$t\mapsto q_t(ix)$$ is a stretched $$\tan$$ with period $$\frac{\pi}{\arctan(x)}$$. In the provided picture with $$c=-0.5$$, $$s=i\sqrt{1/2}$$, the period would be $$\frac{\pi}{\arctan\left(\sqrt{1/2}\right)} \approx$$ 5.104 which matches the visual. bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 08/06/2022, 09:00 AM (This post was last modified: 08/06/2022, 09:02 AM by bo198214.) (08/05/2022, 01:20 AM)JmsNxn Wrote: Also, I think this would only qualify as a local iteration if we allowed poles. Where then we are talking about functions on $$\widehat{\mathbb{C}}$$. So a local iteration about two points independent of $$t$$ will probably be possible, but once extended to its maximal domain will have poles. Yes, but what I meant, it is a local iteration about *one* fixed point. But, yes, there will never be a domain independent of t that contains *both* fixed points - and this is what I announced the picture for in that post. I call it "pole rush": It shows how periodically the pole rushes through the real line, with increasing t (for c=-0.5).     (08/05/2022, 01:20 AM)JmsNxn Wrote: Also, I don't think you even need to put in as much work as you've done to find a counter example. Taking conjugations of $$\lambda^t z$$ by linear fractional transformations, will probably already supply that.Its just that, isn't it? But a very specific one, that shows parabolic plosion on the real axis. JmsNxn Ultimate Fellow Posts: 985 Threads: 117 Joined: Dec 2010 08/07/2022, 10:48 AM (This post was last modified: 08/07/2022, 10:55 AM by JmsNxn.) This thread has become genuine beautiful mercedes, BMW, audi level engineering, Bo. Absolutely fucking beautiful. I don't have any questions. Everything here is just humming like you'd expect german engineering to hum. I'm a huge fan of german mathematics, and have always devoured translations of german mathematics. This is just beautiful, bo. Great fucking additions. You answered all my questions. Ignore any outstanding questions or comments. You just fucking got me thinking so pretty. bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 08/07/2022, 04:41 PM (08/07/2022, 10:48 AM)JmsNxn Wrote: This thread has become genuine beautiful mercedes, BMW, audi level engineering, Bo. « Next Oldest | Next Newest »

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