# Tag Info

2

Using only basic tools of complex analysis $f(1/z)$ is meromorphic at $0$: $\sum_{n=0}^N a_n(z)f(z)^n=0$ with $a_N\ne 0$, take $k$ such that $m=v_0(a_N(1/z)z^{-kN}) \le v_0(a_n(1/z)z^{-kn})$ (order of zero at $0$, negative if a pole) then $\sum_{n=0}^N b_n(z) (z^k f(1/z))^n = 0$ with the $b_n(z)=a_n(1/z)z^{-m-kn}$ holomorphic at $0$ and $b_N(0)\ne 0$. This ...

5

The only sufficient criterion I can think of is the case when $H$ is cyclic of prime power order. Then $Fix(H)$ is a smooth and complex (this holds for any compact group) but also connected by Smith theory. Edit I just checked that Smith theory actually works for arbitrary finite $p$-groups.

3

I doubt this bound can be improved much, at least for even $n$. Indeed: set $z_1 = z_2 = \ldots = z_k = k^{-1/2}$ and $z_{k+1} = z_{k+2} = \ldots = 0$, so that the $\ell^2$ norm of $(z_n)$ is $1$. (Intuitively, this is the worst-case scenario.) Then $$g(\mu) = (E_1(k^{-1/2} \mu))^k = (1 - k^{-1/2} z)^k e^{z \sqrt k} .$$ As $k \to \infty$, the above ...

10

This function is (on the real line, at least) the product of $$\exp( \mu^2 \sum_{i=1}^\infty |z_i|^2 - 2 \mu \Re(\sum_{i=1}^\infty z_i)) \quad (1)$$ and the Hadamard type product $$\prod_{i=1}^\infty E_1( 2 \mu\Re z_i - \mu^2 |z_i|^2) \quad(2)$$ where $E_1$ is the first elementary factor $$E_1(z) := (1-z) \exp(z).$$ The expression (1) is clearly entire in ...

12

The optimal exponent is $k$. Such examples are given by sparse power series. This is actually trivial in the case $k=0$ (which was not included in the OP). Then we can simply take $f(z)=\sum j^{-2} z^{N(j)}$, say. This is obviously bounded, and the coefficients $a_n$ will not satisfy $|a_n|\lesssim n^{-\epsilon}$ for any $\epsilon>0$ if $N(j)$ increases ...

6

To find $f(1)$ to high precision, we will expand $f$ as a Laurent series in $(z+1)^{-1}$, and solve for the coefficients. Setting $f(z)=g(z+1)$, we want to find $$g(z)=z+0+a_1z^{-1}+a_2z^{-2}+\cdots$$ satisfying $$g(z)=z-1+\frac{g(2z-1)}{g(2z)},$$ or $(g(z)-z+1)g(2z)=g(2z-1)$. Hence $$\begin{array}{rcrcrcrcrl} \bigg(1&+&a_1z^{-1}&+&a_2z^{-2}&... 13 Here is a more elementary proof. Suppose F(z,f(z))=0 where F is a polynomial in two variables. How many solutions can the equation f(z)=a for generic a have? Pugging f(z)=a we obtain F(z,a)=0 which has at most d=\deg F solutions. So all equations f(z)=a have at most d solutions, therefore f is rational. I was asked in the comment to stay ... 5 Rohrlich has conjectured that the multiplicative relations in \mathbb{C}^\times / \overline{\mathbb{Q}}^\times between values of \Gamma at rational numbers are generated by the multiplication formula and the reflection formula. In conceptual terms, Lang says that \Gamma is an odd punctured distribution on \mathbb{Q}/\mathbb{Z}, and that conjecturally,... 16 A more abstract argument is also possible: f satisfies p(z,f(z))=0, and let's for convenience assume that p is irreducible (but the argument works in general). We have two meromorphic maps on the associated Riemann surface R=\{ (z,w): p(z,w)=0\}: the standard map (z,w)\mapsto w and also (z,w)\mapsto f(z), this being the composition of (z,w)\... 14 The following argument is based on Christian Remling's proof (given in a comment), but is more elementary. Let us examine the behavior of f(1/z) as z\to 0. The function f(1/z) is algebraic over \mathbb{C}(z), hence there are complex polynomials p_n(z) such that$$\sum_{n=0}^N p_n(z)f(1/z)^n=0.$$Here N is a positive integer. Without loss of ... 3 It is easy to prove that for every region D there exists a function f analytic in D such that \partial D is the "natural boundary" that is f does not have an analytic continuation into any larger region. So whatever you mean by \partial D is 2-dimensional'', you can always have such an example. Such function is easy to construct: ... 4 The integral operator$$Ph(z)=-\frac{1}{\pi}\int\int h(\zeta)\left(\frac{1}{\zeta-z}-\frac{1}{z}\right)dxdy$$acts on L^p, p>2, and the result satiafies (Ph)_{\overline{z}}=h in the sense of distributions. For continuous h, this equation may not have a classical C^1 solution. Edit. The following example was suggested by user @Fedja. It is known ... 4 Existence. (Maybe not a "nice example" as requested, though.) Take any simple closed curve S with Hausdorff dimension 2. (I am assuming you mean Hausdorff dimension when you say "dimension".) Take any function F(z) on the unit disk with the unit circle T as its natural boundary. By the Riemann mapping theorem, we get a ... 1 For a small \alpha > 0, write$$\beta = \frac{\sin^2 \alpha}{\sin(2 \alpha)} = \frac{\tan \alpha}{2} ,$$and define$$g(z) = f(z) \exp(i \beta z^2).$$Then$$|g(z)| \leqslant |f(z)| \leqslant C \exp(|z|^2)$$for z \in A,$$|g(i r)| = |f(i r)| \leqslant M$$for r > 0, and$$\begin{aligned}|g(r e^{i \alpha})| & = |f(r e^{i \alpha}))| \exp(-\...

4

In general, holomorphic maps $f: \mathbb{C} \rightarrow \mathbb{C}$ have no fixed points; but using Picard theorem we can show that $f\circ f$ always have fixed point. Theorem (Fixed-point theorem) Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be holomorphic. Then $f \circ f: \mathbb{C} \rightarrow \mathbb{C}$ always has a fixed point unless $f$ is a ...

7

The set of all polynomials associated with a given one is described as follows: Let the given polynomial be $$P(z)=c(z-z_1)\ldots(z-z_n).$$ Then any associated polynomial is of the form $$Q(z)=\lambda c(z-\sigma_1(z_1))\ldots(z-\sigma_n(z_n)),$$ where each $\sigma_j(z)=z$ or $-\overline{z}$, and $|\lambda|=1$. So, besides the continuous parameter $\lambda$ ...

8

Yes, there is a classification. An isolated branch point can be algebraic or logarithmic. If the branch point is at 0, algebraic means that $f(z^n)$ has a pole or removable singularity at 0. It can also have an essential singularity, but this does not have an accepted name. In the case of a logarithmic point $f(e^z)$ is an (arbitrary) meromorphic function in ...

8

The answer is no in general: for example, if $\phi$ is a cuspidal automorphic form in a cuspidal automorphic representation $\pi$ of $\mathrm{GL}_n(\mathbb{A}_F)$ and $\Phi$ is an Eisenstein series in the noncuspidal automorphic representation $\Pi = \widetilde{\pi} \boxplus \omega$ of $\mathrm{GL}_n(\mathbb{A}_F)$, where $\omega$ is a Hecke character, then ...

12

A differential equation for ${\cal A} (x)$ can be obtained as follows, $$\frac{d^3}{dx^3 } {\cal A} (x) = \int_{-\infty }^{\infty } dk\, (-ik^3 ) e^{ikx} e^{-k^4 } = \frac{x}{4} \int_{-\infty }^{\infty } dk\, e^{ikx} e^{-k^4 } = \frac{x}{4} {\cal A} (x)$$ where integration by parts has been used in the second equality. The leading asymptotic behavior of ...

Top 50 recent answers are included