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.
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.
