Borel summation, Mellin Transforms, Parabolic iteration JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 09/03/2022, 03:02 AM (This post was last modified: 09/03/2022, 04:13 AM by JmsNxn.) So, this will probably be a long series of posts, as I want to get everything out. This post will focus on the general status of representing an Abel function for a parabolic case--and further defining alternative ways of evaluating it. This has been something I've looked at before, but could never crack. Mellin transforms happen pretty naturally for the geometric cases; When the fixed point is attracting. They appear more difficult to do in the parabolic case--require a lot more work. So to begin, we'll get some concepts clear. A parabolic fixed point $$f(0) = 0$$ and $$f'(0) = e^{2 \pi i m/q}$$ for $$q,m \in \mathbb{Z}$$. And the multiplier of this fixed point, as Milnor states it, or the $$\ell = \text{valit}$$ as Bo, refers to it, is given in the expansion: $$f^{\circ q}(z) = z \left(1 + az^\ell + o(z^\ell)\right)\\$$ Now, working with $$f$$ is largely unnecessary; so we can make the change of variables $$f = f^{\circ q}$$, so that we just focus on when $$f'(0) = 1$$. A standard result, is that now we have $$2 \ell$$ petals about $$0$$, where there are $$\ell$$ attracting petals, and $$\ell$$ repelling petals. For clarity--these petals can be topologically identified as follows. Let $$P$$ refer to one of these $$2\ell$$ petals--then the union $$\mathcal{N} = \bigcup P$$ is a neighborhood of $$0$$ itself.  The smallest difference is that, for attracting petals, we have that $$f(P) \subset P$$--where $$f^{\circ n}(P) \to 0$$. And for repelling petals, we have the inverse relation: $$f^{-1}(P) \subset P$$--where $$f^{\circ -n}(P) \to 0$$. Now an interesting phenomenon is that we get a star shaped domain about $$0$$, if we discuss $$|z| < \delta$$ for $$\delta$$ small enough. We will have $$2 \ell$$ vectors, perfectly placed around $$0$$--with the $$2 \ell$$ roots of unity. And these vectors describe perfectly the boundaries between parabolic iteration at 0. Now the value $$a$$ in the above expansion is unneeded, it only serves to perturb the following conjugation. Let's choose a vector $$\mathbf{v}$$ in either of these above directions. Let's write: $$\varphi(z) = \frac{\mathbf{v}}{a\sqrt[\ell]{z}}\\$$ And let's write: $$F(w) = \varphi^{-1} \bullet f \bullet \varphi \bullet w\\$$ Where, these bullets are just short hand for composition. So it looks like $$F(w) = \varphi^{-1}(f(\varphi(w)))$$. This changes the domain of interest into $$\mathcal{H}_R$$ where: $$\mathcal{H}_R = \{w \in \mathbb{C}\,|\,\Re(\mathbf{v}w) > R\}\\$$ Where now, $$F : \mathcal{H}_R \to \mathcal{H}_R$$--and $$F$$ looks like $$w \mapsto w+1$$. And since this is a half plane with a fixed point at infinity, we can change this into just a standard half plane (Through the conjugation $$F(w+R)-R = F$$): $$F(w) : \mathbb{C}_{\Re(w) > 0} \to \mathbb{C}_{\Re(w) > 0}\\$$ And on this reparameterized half plane, we have the expansion: $$\alpha(w) : \mathbb{C}_{\Re(w) > 0} \to \mathbb{C}_{\Re(w) > 0}\\$$ Where, $$\alpha(F(w)) = \alpha(w) + 1$$. -------------------------------------------------------- Here is where things start to get difficult. We have an abel function on half plane, which is really just the abel function of our function $$f$$, but fixated in a petal, and near $$0$$. We have to from here, restrict (we don't have to in the final result, but we do now). This function $$\alpha(w)$$ satisfies the asymptotic expansion: $$\alpha(w) = w + A \log(w) + \sum_{k=0}^\infty b_k w^{-k}\\$$ Where the coefficients $$b_k$$ are of order $$O(c^kk!)$$ for some random value $$c$$ which is unimportant to us. This asymptotic only holds for $$|\arg(w)| < \kappa$$, for some value $$\kappa>0$$--dependent on the multiplicity of this fixed point (Milnor), or the valit, as Bo calls it. And so enters in our new change of variables $$u = 1/w$$: $$\alpha(u) = \frac{1}{u} - A \log(u) + \sum_{k=0}^\infty b_k u^k\\$$ And for $$G(u) = 1/F(1/w)$$--which is holomorphic in the half plane $$\Re(u) > 0$$, with a singularity at $$u=0$$ and bounded behaviour at $$u = \infty$$. Then: $$\alpha(G(u)) = \alpha(u) + 1\\$$ We are finally in the closed form we need to start talking about Mellin transforms. ------------------------------------------------------------- Let's write: $$H(s) = \int_0^\infty \alpha(u)u^{s-1}\,du\\$$ And forego, momentarily where this converges. We first split this integral into $$\int_0^1 + \int_1^\infty$$. Then, the first integral can be expanded as: $$\int_0^1\left(w^{s-2} + w^{s-1}\log(w)\right)\,dw + \sum_{k=0}^\infty \frac{b_k}{s+k}\\$$ This series does not converge, but it doesn't matter, as it's a formal expression constructed from an analytic function $$\int_0^1 \alpha(u) u^{s-1}\,du$$. The second integral: $$\int_1^\infty \alpha(u)u^{s-1}\,du\\$$ Converges for all $$\Re(s) < 0$$, because $$|\alpha(u)| < M$$ for all $$\Re(u) > 1$$. This tells us additionally that: $$H(s) = e^{ -\frac{\pi}{2} |\Im(s)|}\\$$ Which is just a Stein and Shakarchi argument on holomorphic Fourier transforms. (This integral exists in a half plane, therefore must have this decay). But additionally, this also tells us that $$H(s)$$ is analytically continuable to $$\Re(s) < 0$$, and has this decay through out. ---------------------------------------------------------------------- Now let's blow shit up, since this is invertible, we have that: $$\alpha(u) = \int_{c-i\infty}^{c+i\infty}H(s)u^{-s}\,ds\\$$ For $$c \approx 0$$, works perfectly for all $$u$$ in the right half plane. Don't appoint me for this result; this is Stein and Shakarchi's Fourier inversion theorem, but applied for the mellin transform. It's found blatantly in D. Zagier's paper I posted before. I have written it somewhat uniquely, but it is backable by  results which have nothing to do with my own work And now, we can represent the abel function using a borel sum. If I write: $$T_c(u) = \int_{c-i\infty}^{c+i\infty} \frac{H(s)}{\Gamma(1-s)} u^{-s}\,ds\\$$ We get the perfect version of: $$T_c(u) = \sum_{k=1}^\infty \frac{b_k}{k!}u^k$$ Where we can choose the index $$k=1$$ by moving $$c$$ to the right or left. For $$c = \delta - N$$ we get: $$T_c(u) = \sum_{k=N}^\infty \frac{b_k}{k!}u^k$$ Where, (I'm not sure if this converges, if it doesn't, it's irrelevant), if $$c = 1+\delta$$ we have the borel sum of the Alpha function. Therefore: $$\int_0^\infty e^{-t/u}T_{1+\delta}(t) \,dt = \alpha(u)\\$$ But now, in addition to everything above, this will prove A GLOBAL SOLUTION OF OUR BOREL SUM METHODOLOGY! ----------------------------------------------- I still have to iron out a lot of details; but this is a much more effective approach at dealing with global solutions. So that we can write our borel sum on a larger domain than the naive borel sum gets us. I definitely made a few typos here. I'm trying to build a large result, but I thought I should share what I have so far . I'm taking all questions, I want to do this much better than done here. But I truly believe I can make the argument that Fractional Calculus can be applied to parabolic iteration--which has been a dream of mine for awhile, lol. Regards, James -------------------------------------------------- Next up, I'll explain how to achieve fractional iterates using the mellin transform, and this new Borel sum knowledge. This is solely for the abel function--the fractional iteration gets much more interesting. Especially as we relate it to the half plane--the change of variables required is difficult. bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 09/09/2022, 08:23 PM (This post was last modified: 09/09/2022, 08:55 PM by bo198214.) (09/03/2022, 03:02 AM)JmsNxn Wrote: So to begin, we'll get some concepts clear. A parabolic fixed point $$f(0) = 0$$ and $$f'(0) = e^{2 \pi i m/q}$$ for $$q,m \in \mathbb{Z}$$. And the multiplier of this fixed point, as Milnor states it, or the $$\ell = \text{valit}$$ as Bo, refers to it, is given in the expansion:I think you mean "multiplicity", perhaps edit that in your original post - because it is called "multiplier" in the hyperbolic case for the first coefficient, right? EDIT: After reading Milnor again: the multiplicity is valit+1, they are not the same. The multiplicity is the index of the first non-zero coefficient after the first coefficient. Valit is the multiplicity - 1. The number of petals is 2*valit. (09/03/2022, 03:02 AM)JmsNxn Wrote: Here is where things start to get difficult. We have an abel function on half plane, which is really just the abel function of our function $$f$$, but fixated in a petal, and near $$0$$. We have to from here, restrict (we don't have to in the final result, but we do now). This function $$\alpha(w)$$ satisfies the asymptotic expansion: $$\alpha(w) = w + A \log(w) + \sum_{k=0}^\infty b_k w^{-k}\\$$ Where the coefficients $$b_k$$ are of order $$O(c^kk!)$$ for some random value $$c$$ which is unimportant to us. This asymptotic only holds for $$|\arg(w)| < \kappa$$, for some value $$\kappa>0$$--dependent on the multiplicity of this fixed point (Milnor), or the valit, as Bo calls it. But I would like to remark here that asymptotics is for all petals the same, only the branch(cut) of the logarithm is what needs to be adjusted from petal to petal. (09/03/2022, 03:02 AM)JmsNxn Wrote: And so enters in our new change of variables $$u = 1/w$$: $$\alpha(u) = \frac{1}{u} - A \log(u) + \sum_{k=0}^\infty b_k u^k\\$$Don't you need more terms $$\frac{1}{u^k}$$ and also having coefficients, like: $$\alpha(u) = \left(\sum_{k=1}^{\ell} \frac{c_k}{u^k}\right) - A \log(u) + \sum_{k=0}^\infty b_k u^k$$ (09/03/2022, 03:02 AM)JmsNxn Wrote: $$T_c(u) = \int_{c-i\infty}^{c+i\infty} \frac{H(s)}{\Gamma(1-s)} u^{-s}\,ds\\$$ We get the perfect version of: $$T_c(u) = \sum_{k=1}^\infty \frac{b_k}{k!}u^k$$ Where we can choose the index $$k=1$$ by moving $$c$$ to the right or left. For $$c = \delta - N$$ we get: $$T_c(u) = \sum_{k=N}^\infty \frac{b_k}{k!}u^k$$ Where, (I'm not sure if this converges, if it doesn't, it's irrelevant), if $$c = 1+\delta$$ we have the borel sum of the Alpha function. Therefore: $$\int_0^\infty e^{-t/u}T_{1+\delta}(t) \,dt = \alpha(u)\\$$ But now, in addition to everything above, this will prove A GLOBAL SOLUTION OF OUR BOREL SUM METHODOLOGY!Where do $$\delta$$ and N come from? JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 09/10/2022, 05:13 AM (This post was last modified: 09/10/2022, 05:16 AM by JmsNxn.) (09/09/2022, 08:23 PM)bo198214 Wrote: (09/03/2022, 03:02 AM)JmsNxn Wrote: So to begin, we'll get some concepts clear. A parabolic fixed point $$f(0) = 0$$ and $$f'(0) = e^{2 \pi i m/q}$$ for $$q,m \in \mathbb{Z}$$. And the multiplier of this fixed point, as Milnor states it, or the $$\ell = \text{valit}$$ as Bo, refers to it, is given in the expansion:I think you mean "multiplicity", perhaps edit that in your original post - because it is called "multiplier" in the hyperbolic case for the first coefficient, right? EDIT: After reading Milnor again: the multiplicity is valit+1, they are not the same. The multiplicity is the index of the first non-zero coefficient after the first coefficient. Valit is the multiplicity - 1. The number of petals is 2*valit. Oh yes, sorry. I misplaced my terms. What I meant is still the same as you though, I just thought the multiplicity = valit. Either way I meant that: $$f(z) = z(1+az^n + o(z^n))\\$$ And that $$n =$$ valit. I guess I just changed the variables to fast, and forgot the nomenclature that the multiplicity (not multiplier--I'm bad with language, bo) is $$n+1$$, you are right. Quote:But I would like to remark here that asymptotics is for all petals the same, only the branch(cut) of the logarithm is what needs to be adjusted from petal to petal. Yes, you are absolutely correct, I had not noticed that entirely yet. I thought there was something more subtle going on, but yes. You are correct. I was also just trying to be reserved in my estimations of the result. Quote:Don't you need more terms $$\frac{1}{u^k}$$ and also having coefficients, like: $$\alpha(u) = \left(\sum_{k=1}^{\ell} \frac{c_k}{u^k}\right) - A \log(u) + \sum_{k=0}^\infty b_k u^k$$ OKAY! So you are right here to a point--but not exactly. I've been working a lot lately, so I haven't had time to explain these constructions perfectly. I think to draw out the mellin transform isomorphism, is the right move here--and further describes what I am getting at. Here is where things get spicy! Bo, I just want you to remember we are doing all of these variable changes, "under the integral". So When I write: $$g(w) \sim w + \log(w) + \sum_{k=0}^\infty b_k w^{-k}\\\$$ And we know that: $$\int_0^\infty g(w)w^{s-1}\,dw\\$$ Converges, we can invert, as such, changing $$u = 1/w$$ and $$du = -\frac{dw}{w^2}$$ creates an interchange under the integral: $$\int_0^\infty g(1/u)u^{-s-1}\,du\\$$ And now the asymptotic series is at $$0$$--rather than at infinity. And we can do the above transformation/asymptotic expansion. I believe your mistake is thinking we have to conjugate the Abel function--which we don't do. We conjugate the function the abel function is associated with--within the Abel function. By which, I mean: $$\alpha(u) = \alpha(1/w)\\$$ We do not perform a flip on the series itself, just on the inner terms. So I believe you are mistaken here. At least, as I know of it. I'm going to draw this out a bit, just to paint a more vivid picture. Assume that: $$\alpha(f(w)) = \alpha(w) + 1\\$$ And now let's define a function $$F$$ such that $$F(w) = \frac{1}{f(1/w)}$$. Then, the function $$A(w) = \alpha(1/w)$$, satisfies: $$A(F(w)) = A(w) + 1\\$$ Which, I'm sure you can check. Quote:Where do $$\delta$$ and N come from? Oh, I apologize here, Bo. These are terms to do with Mellin transforms--and are essentially variables which act as limiters. The value $$N$$ is a natural number greater than $$0$$, and each of these expansions are valid for all $$N$$--and $$\delta>0$$, is my way of saying a small value $$\delta$$. Where such a value always exists, you may just need to choose small enough. I think I need to write a much more in depth construction of these concepts, I can kind of just blaze through it, and forget you guys aren't as familiar with Mellin transform/analytic number theory kind of stuff. I will definitely try my best to explain everything though, bo. Apologize if this is confusing. I really want us to shift from "Borel Sums" into, this exists in a Mellin transformed space and we transform back. Which is the "sophisticated" way of doing borel sums, lol. D. Zagier's paper is sort of my bible right now. I'm working on writing a much better write up, I'm just not sure if I have the time to write a full paper on this. But Parabolic iteration is making so much fucking sense that it never made before--precisely because I can do this Mellin shit now. Which, I guess, is just how my mathematics works. I'll work harder at explaining the details  Regards, James bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 09/10/2022, 06:52 AM (09/10/2022, 05:13 AM)JmsNxn Wrote: Oh yes, sorry. I misplaced my terms. What I meant is still the same as you though, I just thought the multiplicity = valit. Either way I meant that: $$f(z) = z(1+az^n + o(z^n))\\$$ Yes, no biggie. Just wanted us to be clear about the terms. (09/10/2022, 05:13 AM)JmsNxn Wrote: Quote:Don't you need more terms $$\frac{1}{u^k}$$ and also having coefficients, like: $$\alpha(u) = \left(\sum_{k=1}^{\ell} \frac{c_k}{u^k}\right) - A \log(u) + \sum_{k=0}^\infty b_k u^k$$ I believe your mistake is thinking we have to conjugate the Abel function--which we don't do. We conjugate the function the abel function is associated with--within the Abel function. By which, I mean: No, actually I didnt care about the conjugations, I were just relying on (09/03/2022, 03:02 AM)JmsNxn Wrote: We have an abel function on half plane, which is really just the abel function of our function $$f$$, but fixated in a petal, and near $$0$$. and the Abel function of $$f$$ looks like I described above ... but with all those conjugations ... I mean you first conjugate with $$\varphi(z) = \frac{\mathbf{v}}{a\sqrt[\ell]{z}}$$ but then you conjugate with $$1/z$$ which would be imho the conjugation $$\frac{a\sqrt[\ell]{z}}{\mathbf{v}}$$ ? Something does not yet feel right about the Abel formulas (apart from that they don't converge of course.) JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 09/10/2022, 08:00 AM (This post was last modified: 09/10/2022, 08:57 AM by JmsNxn.) I might be making a mistake somewhere. I know the mellin transforms don't necessarily converge; but they necessarily produce a mellin transformed function. This sort of goes "distributional theory", but not totally. We get that--$$f(0) = 0$$, and $$f'(0) = 1$$, valit 1. There by we can rewrite this $$F(w) = 1/f(1/w)$$. Now from here we have a half plane $$H_R$$ for $$\Re(w) > R$$, where $$F(w) \sim w + 1 + o(w)$$, whereby $$F$$ is injective on $$H_R$$. Now let's do the conjugation $$G(w) = F(w+R) - R$$. Now, $$G$$ is holomorphic on a right half plane $$\Re(w) > 0$$. What I had produced as my conclusions, were based on this. From here, we can make another switch $$G(u) = G(1/w)$$, which fully produces our expansion at infinity if $$B(G(u)) = B(u) + 1$$, for a function $$B$$. And $$B$$ is the same thing, it is an LFT $$\ell$$ applied to $$\alpha(f)$$. It looks something like $$\alpha(\ell(u)) = B(u)$$. Where $$\ell(G) = f(\ell)$$. Can I ask what you are trying to numerically evaluate, so I can chime in? I don't know how you're approaching it. This would be exceedingly difficult to program in. I can't think of any obvious manner of evaluating this efficiently and effectively. You have to find $$R$$, you have to perform conjugations, and require taylor data at every step. bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 09/10/2022, 03:12 PM (09/10/2022, 08:00 AM)JmsNxn Wrote: Can I ask what you are trying to numerically evaluate, so I can chime in?I don't, but actually I wondered how the stuff you wrote would lead to a numerical procedure. But I am patient - just continue! With the Abel function I just wondered why it looks different than I know it, but we can leave this aside for now. « Next Oldest | Next Newest »

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