18
votes

Accepted

### Can the topologist's sine curve be realized as a Julia set?

The answer is negative. Since every neighborhood of a point
on the Julia set is mapped onto the whole Julia set by some iterate of the rational function, it follows that if a small piece of the Julia ...

8
votes

Accepted

### Is the real part of the Eta function bounded by $2 \sum_{n=1}^{\infty}{\dfrac{(-1)^{n-1}}{n^{\alpha}}} $

The question is whether this inequality on the Dirichlet eta function holds:
$$\Re\eta(\alpha+i\beta)\leq 2\eta(\alpha).$$
It does not hold.
Here is a plot of $\Delta(\alpha,\beta)=2\eta(\alpha)-\Re\...

5
votes

Accepted

### Sign of $\Re\frac{\xi'(s)}{\xi(s)}$ locally around a zeta zero

If a complex function $f(s)$ has a zero at $s_0$ of order $n\geq 1$, then for suitable $r>0$,
$$\frac{f'(s)}{f(s)}=\frac{n}{s-s_0}+O(1),\qquad 0<|s-s_0|<r.$$
Writing $s=s_0+\rho e^{it}$ with $...

4
votes

Accepted

### On $\Re\frac{\zeta'}{\zeta}(s) \geq -A \log(|t|+4)$ for some $A>0$

If the Riemann Hypothesis fails, then there is no such lower bound. Indeed, if $\zeta(s_0)=0$ and $\Re(s_0)>1/2$, then
$$\lim_{\delta\to 0+}\Re\frac{\zeta'(s_0-\delta)}{\zeta(s_0-\delta)}=-\infty.$$...

3
votes

### Is a $2$-form which is "almost" Kähler cohomologous to a Kähler form?

Such form is called "taming the almost complex structure"
or "Hermitian symplectic". This is a famous problem known as
"Streets-Tian conjecture": are all compact complex ...

3
votes

Accepted

### Estimates for the first coefficient of the Taylor expansion of $\zeta$ around a zero

Since
$$c_1=\zeta'(s_0)\qquad\text{and}\qquad c_1'=\zeta'(1-\overline{s_0})=\overline{\zeta'(1-s_0)},$$
the question concerns $|\zeta'(s_0)/\zeta'(1-s_0)|$. The functional equation for $\zeta(s)$ can ...

2
votes

Accepted

### Generalisation of Paleyâ€“Wiener type results for unbounded sets

I am not sure if this is the sort of thing you want, but here it goes anyway.
The idea is that if $f\in L^2(\mathbb{R}^n)$ and $A$ is very thin then by the Cauchy--Schwarz inequality $f$ is an entire ...

2
votes

### Motivation for defining polar derivative

The motivation (and the word "polar") comes from algebraic geometry. Consider the homogenized polynomial $P(x,y)=y^n p(x/y)$, where $p$ is a complex polynomial of degree $n$. The polar of $P$...

2
votes

Accepted

### Riemann xi function strictly increasing along a half-plane

The result you mention is not due to Matiyasevich-Saidak-Zvengrowsk. Instead, it appeared in Sondow-Dumitrescu: A monotonicity property of Riemann's xi function and a reformulation of the Riemann ...

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