Skip to main content
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 ...
Alexandre Eremenko's user avatar
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\...
Carlo Beenakker's user avatar
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 $...
GH from MO's user avatar
  • 101k
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 $$\lim_{\delta\to 0+}\Re\frac{\zeta'(s_0-\delta)}{\zeta(s_0-\delta)}=-\infty.$$...
GH from MO's user avatar
  • 101k
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 ...
Misha Verbitsky's user avatar
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 ...
GH from MO's user avatar
  • 101k
2 votes
Accepted

Generalisation of Paley–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 ...
Aleksei Kulikov's user avatar
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 $P$...
N M's user avatar
  • 306
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 ...
GH from MO's user avatar
  • 101k

Only top scored, non community-wiki answers of a minimum length are eligible