# Tag Info

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 ...
• 89.8k
Accepted

• 101k
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 $$\lim_{\delta\to 0+}\Re\frac{\zeta'(s_0-\delta)}{\zeta(s_0-\delta)}=-\infty.$$...
• 101k

### 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 ...
• 9,145
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 ...
• 101k
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 ...
• 5,506

### 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 $P$...
• 306