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1 vote

What is the relationship between $\sum_{n=0}^\infty f(n) x^n$ and $-\sum_{n=1}^\infty f(-n) x^{-n}$?

Some extended comments: Investigations of this type appear in the theory of the Mellin transform and, in particular, the use of Ramanujan's master heuristic (theorem, formula) as illustrated in the MO-...
  • 8,551
4 votes

What is the relationship between $\sum_{n=0}^\infty f(n) x^n$ and $-\sum_{n=1}^\infty f(-n) x^{-n}$?

This is also just a comment but large: It might be worthwhile to try poking around with this formula on functions which cannot be analytically continued to see what arises. Trying to break the formula ...
8 votes

What is the relationship between $\sum_{n=0}^\infty f(n) x^n$ and $-\sum_{n=1}^\infty f(-n) x^{-n}$?

First of all, to make sense of $f(-n)$ we need some assumptions about $f$. For example, let $$\sum_{n=0}^\infty f(n)z^n\quad\quad\quad\quad (1)$$ be a series with positive radius of convergence. Then ...
6 votes

What is the relationship between $\sum_{n=0}^\infty f(n) x^n$ and $-\sum_{n=1}^\infty f(-n) x^{-n}$?

I'll convert my comment to an answer. One issue that you're implicitly sweeping under the rug is that we need to be able to make sense of the evaluation of $f$ at negative integers. Given an arbitrary ...
  • 19.9k
7 votes

What is the relationship between $\sum_{n=0}^\infty f(n) x^n$ and $-\sum_{n=1}^\infty f(-n) x^{-n}$?

This is just a comment but I am not entitled and it will be too long anyway. If we start with a toy example where $f(n)=1$ for all $n$, then $F$ is the function $\frac 1{1-z}$ in the open disc, $...
  • 181
7 votes

What, exactly, has Louis de Branges proved about the Riemann Hypothesis?

In the years since this question was first asked and answered, there have been some new developments (namely, 80+ pages of commentary by Eric Kvaalen, and a large number of new versions of de Branges' ...
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3 votes
Accepted

Is a local isometry of the hyperbolic plane the restriction of a global isometry?

The hyperbolic plane has this property, as does the Euclidean plane. If $E$ is any subset of $ \mathbb{H}^2$, and $u : K \to\mathbb{H}^2$ is an isometry, then there is an extension of $u$ which is an ...
  • 38.5k
2 votes

Proper journal for a preprint in complex geometry

As far as I know, if not already peer-reviewed and then published in a journal, the manuscript that you are going to submit for peer-review is still called a preprint (even if it has already been ...
  • 313
5 votes

Holomorphic maps from a Riemann surface of infinite genus

The following is not a real answer but an extensive comment on the OP. First, recall that an (open) Riemann surface is said to have type $P_{AB}$ (resp. $O_{AB}$) if it admits (resp., does not admit) ...
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4 votes
Accepted

On a lemma of Łojasiewicz in complex analysis of one variable

The assumption of the Lemma is that $f$ has a finite limit as $z\to \zeta_0$. This assumption does not hold in any of the two examples that you mention. In these examples, $f$ has a limit only when $z\...
3 votes

Holomorphic maps from a Riemann surface of infinite genus

Edit: as Moishe points out, my answer (just below) is for a different, and easier, question. I will leave the answer up, as it does feel “related”. The answer is no. I find it conceptually easier to ...
  • 21.1k
3 votes
Accepted

Quantitative analytic continuation estimate for functions small except on a small set

This conjecture is correct. Take $K=e$, and let $\gamma\leq 1/4$; we will fix $\gamma$ later. First we give a crude estimate of $c_0$. Let $g(z)=\sum_{1}^\infty c_nz^n.$ Since $|c_n|\leq e^n$, we ...
4 votes
Accepted

Entire function with almost periodic boundary condition?

The answer seems to be negative. Suppose that an entire function $f$ satisfies $f(z+v_i)=e^{A_iz}f(z)$, where $v_1$ and $v_2$ generate a lattice. Let $\Pi$ be the fundamental parallelogram of this ...
5 votes

Quantitative analytic continuation estimate for a function small on a set of positive measure

$\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$As shown in the post by Alexandre Eremenko, the answer to this question is yes if $C=1$ (and hence if $C\le1$). On the ...
11 votes
Accepted

a problem in complex-variable inequality

For a polynomial $Q(x)=\sum_i q_ix^i$, define $N(Q)=\sum_i|q_i|^2$. We need to show that $N(R)\geq 2$, where $R(x)=\prod_i (x-z_i)$. For a polynomial $Q(x)=\sum_{i=0}^k q_ix^i$, define $Q^*(x)=\sum_{i=...
7 votes

Quantitative analytic continuation estimate for a function small on a set of positive measure

The answer depends on $C$. For example, for $C=1$ it is positive. Your estimate $|f^{(m)}(0)|\leq m!$ implies that $|f_n(z)|\leq 1/(1-|z|).$ Take $|z|=1/2$, you conclude that $|f_n(z)|\leq 2,\; |z|<...
11 votes
Accepted

Quantitative analytic continuation estimate for a function small on a set of positive measure

Unfortunately, no, as requested: Take any sequence $\delta_j\in(0,1)$ decaying to $0$, choose small $\mu_j>0$ such that $\prod_j \delta_j^{\mu_j}=e^{-1}$ and put $f_n(z)=e^n\prod_j B_{\delta_j}(z)^{...
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4 votes
Accepted

Bounds on zeros of rational function

Let us drop the assumption $x_j\in[1,2]$, it is not needed. Proving the result by contradiction, denote our function by $f_N$, suppose that $f_N(z_N)=-i$, and $\mathrm{Im}\ z_N= 1/(N^2R_N)$ where $R_N\...
2 votes

A question on the use of fractional derivatives in Riemann Hypothesis

(Disclaimer I need to verify if your integral of $\frac{1}{1+e^{iw}}$ trick works but here's an alternative framework which should meet your needs) So we wish to evaluate $$(1-2^{1-a} ) \zeta(a) = \...
0 votes

Riemann uniformization theorem (limit case)

I'll attempt to sketch a proof that this is true. First, it is convenient to apply the map $z\mapsto \log z$, which maps annular regions in question to thin $2\pi$ - periodic vertical strips $S_r$ and ...
  • 6,274
6 votes

An inf-sup estimate for holomorphic functions

This is not true: take $n=1$, $r=1$, $\eta(z)=e^{az},\; a>0,$ then $$\max_{z\in B(0,r)}|\eta(z)|=e^a,$$ while $$\min_{z\in B(0,1)}|\eta(z)|=e^{-a}.$$ Since $a>0$ is arbitrary, no $\kappa$ with ...

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