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Extending the argument by GH from MO, $\zeta(P(s))$ has a pole for any $s$ such that $P(s)=1$, while $P(\zeta(s))$ has unique pole for $s=1$. Therefore if $\zeta(P(s))=P(\zeta(s))$, then $P(s)=1$ has unique solution $s=1$ and $P(x)-1=c(x-1)^n$ for some complex $c$ and positive integer $n$, and we get $\zeta(1+c(s-1)^n)=1+c(\zeta(s)-1)^n$. For $s=1+x$ with ...
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It doesn't seem to me that either article follows the line of reasoning as you have presented it. Indeed, we do not take the integral over the union of integrals $(x_n,(1+\varepsilon)x_n)$. We do get that the integral over those intervals is infinite, but you correctly note this does not give a contradiction. Instead, the argument goes as follows.
Let me ...
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Main areas are dynamics of automorphisms (for example, Henon maps), dynamics of endomorphisms, dynamics of foliations, and local dynamics.
Eric Bedford, Tien-Cuong Dinh, John Fornaess, Misha Lyubich, Nessim Sibony, John Smiley, and their students and collaborators.
One-dimensional holomorphic dynamics, functions of several complex variables (especially ...
9
It is straightforward to see that if $P\in\mathbb{C}[x]$ satisfies $P(\zeta(s))=\zeta(P(s))$, then $P(1)=1$. Note that if $P(\zeta(s))=\zeta(P(s))$ holds for all real $s>1$, then it also holds for all complex $s\neq 1$ by the uniqueness of analytic continuation.
Indeed, $\zeta(s)$ has a pole at $s=1$, hence $P(\zeta(s))=\zeta(P(s))$ also has a pole at $s=...
9
No. If $P$ has a positive leading coefficient, then letting $s$ go to infinity and using continuity of $P$ we get $P(1)=1$. If $P$ has a negative leading coefficient, then $\zeta(P(s))$ has zeros at arbitrarily large $s$, while $P(\zeta(s))$ does not.
6
Here is a simple argument which applies to polynomials in $C[x]$, and even to most rational functions in $C(x)$. $\zeta$ is a meromorphic function in the plane.
So the Nevanlinna characteristic $T(r,\zeta)$ is defined. It is a positive increasing function, a kind of transcendental analog of the degree of a rational function. Your equation implies
that $(\deg ...
6
The identity
$$\sum_{n=1}^\infty\frac{\mu(n)\log n}{n}=-1$$
was conjectured by Möbius (1832) and proved by Landau (1899). It is a consequence of the prime number theorem. Not surprisingly, the rate of convergence is determined by the (known) zero-free region of $\zeta(s)$. In particular,
$$S(x)=-1+O_\epsilon(x^{\alpha-1+\epsilon})$$
holds for any $\epsilon&...
4
Let $z_0\in\Bbb{R}$ be arbitrary. The matrices $A_0:=A(z_0)$ and $B_0:=B(z_0)$ are normal; in view of $A_0^*=A^\#(\bar{z_0})=A^\#(z_0)$ and $B_0^*=B^\#(\bar{z_0})=B^\#(z_0)$ they commute with their Hermitian adjoints. They also commute with each other. So $A_0$ and $B_0$ could be simultaneously diagonalized by a unitary matrix. That matrix also diagonalizes $...
4
A partial solution: As mentioned by @MargaretFriedland, the desired $M_a(r)$ is the absolute maximum of $g(t):=\sqrt{{\rm{e}}^{2r\cos t}+2a{\rm{e}}^{r\cos t}\cos(r\sin t)+a^2}$. Notice that this is an even function, so it suffices to maximize over $[0,\pi]$. If $t\in\left[\frac{\pi}{2},\pi\right]$, then $\cos t<0$. So ${\rm{e}}^{2r\cos t}\in [0,1]$ and, ...
4
An answer before the numerics start: First, note that for $w=u+iv \in \mathbb{C}$ and a fixed $a \in \mathbb{R}$ we have $|w+a|=\sqrt{u^2+v^2+2au+a^2}=\sqrt{|w|^2 + 2au+a^2}$ . Next, consider the image under the complex exponential of the circle $|z|=r>0$. Using polar coordinates, we get $|\exp (r\cos t+i r\sin t)|=e^{r\cos t}$. Thus to get $M_a(r)$ we ...
4
You should be able to find anwers to your question in a well witten book by Franc Forstnerič: Stein Manifolds and Holomorphic Mappings -The Homotopy Principle in Complex Analysis, https://www.springer.com/gp/book/9783319610573
In particular look at sections 9.10, 9.11 and Theorem 9.11.5. The book contains all the background material you need and also ...
4
The unique measure of maximal entropy $\mu_f$ supported on the Julia set of a rational map $f$ of degree $d \geq 2$ is indeed the unique balanced measure for $f$, i.e., the only probability measure $\mu$ not charging the exceptional set and satisfying $f^*\mu =d \cdot \mu$. As you already noticed in the comments, uniqueness of a measure with this property ...
2
The measure of maximal entropy is the unique measure that is "fully invariant" in your sense. I believe that this already follows from the original proofs - indeed, it is well-known that if you take a point mass at some non-exceptional point, and keep pulling back, you will converge to the measure of maximal entropy. This should be enough to deduce the claim....
1
I think this an issue of choice of branch of square root. Note that the paper asks you to use the branch holomorphic in $\mathbb{C} \setminus E$ and behaves as $z$ as $z \to \infty$.
Let's call this choice of branch $w(z)$ and call the compact set delimited by $[-1, 1] \cup E$ by $K$. Then the map we are looking for can be defined by
$$ z + w(z) = \left\{ ...
1
Assuming that a closed formula is not accessible, possibly maxmod, which calculates the maximum modulus of a complex polynomial on the unit disk, would speed a numerical attack. (Polynomial order of a few hundred is not a problem.)
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The way that the paper by Freire-Lopes-Mañé makes sense of $f^*\mu=d\mu$ is the following: ''For any Borel subset $A$ of $\Bbb{C}_\infty$ with $f\restriction_A$ injective, one has $\mu(f(A))=d.\mu(A)$.'' (See p. 46 of this paper.)
One observation is that such an ergodic measure $\mu$ is either supported on the Julia set or is one of those measures with ...
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