# Tag Info

6

This is not really an answer, it is a comment with images. The paper "Quadrilateral Mesh Generation II : Meromorphic Quartic Differentials and Abel-Jacobi Condition" (arXiv 2019) contains (imo) nice exposition of theoretical background and several illuminating illustrations.

6

The solutions of $(L-\lambda)u=0$ are the functions $u(x)=e^{i\lambda x}v(x)$, where $v$ satisfies $Lv=0$. The periodicity amounts to $e^{i\lambda}v(1)=v(0)$. Thus your problem does admit infinitely many solutions. Just consider the monodromy matrix $M:v(0)\mapsto v(1)$, whose determinant equals $1$ (by the Wronskian). Take an eigenvalue $\mu$ of $M$, which ...

5

Yes, this is true. By the boundedness assumption, $u$ extends to a (bounded) $\omega$-psh function on $X$. Bedford and Taylor defined in '82 the Monge-Ampère operator of a bounded psh function, which has later been extended to the global quasi-psh case. Almost by definition of that operator, you have $\int_X(\omega+dd^c \varphi)^n=\int_X \omega^n$ for any ...

5

There is a very nice hydrodynamic interpretation, going back to Riemann and Klein, and popularized by Courant (Hurwitz and Courant, Function theory). For those who do not read German, there is a modern exposition in French: E. Ghys and D. Smai, Six lecons autour des surfaces de Riemann Edit. Klein's book is "On Riemann's theory of algebraic functions&...

4

Yes, there exists a quasiconformal map (even a homeomorphism) from the Riemann sphere to a polytope. Embedding to $R^3$ is irrelevant here, all we need is the intrinsic metric, which is a flat metric with conic singularities. Consider a conic singularity with angle $2\pi\alpha$. Map a neighborhood of $0$ onto a neighborhood of this point by the map which has ...

4

TL;DR Peter Taylor pointed out that the expressions below reduce to $$\lim_{n\rightarrow k}x^{-n}I_n(x)= (x-1) \sum_{i=0}^k \frac{H_{k+1} - H_i}{x^{i+1}} -2 \,\text{arctanh}\,(1-2 x)$$ $\newcommand\HGF{_2\!\tilde{F}_1}$Mathematica is able to compute these limits, the result is in terms of a partial derivative of the regularized hypergeometric function $\HGF$...

4

The key is the following integral representation: \begin{aligned} s_n(z)&:= \sum_{j>n} \frac{z^j e^{-j/n}}j \\ &=\sum_{j>n} z^j e^{-j/n} \int_0^\infty du\,e^{-ju} \\ &=\int_0^\infty du\,\sum_{j>n} z^j e^{-j/n}e^{-ju} \\ &=\frac{z^{n+1}}{e^{1+1/n}}\,\int_0^\infty du\,\frac{e^{-(n+1)u}}{1-z e^{-u-1/...

3

Great question! So far, we haven't been able to produce analytic ring structures on $\mathbb C$-algebras that are not induced. Similarly, if we equip $\mathbb Q_p$ with a liquid analytic ring structure, we are also only able to use induced analytic ring structures: To "overconvergent rigid spaces"(=Größe-Klönne's dagger spaces), one can associate ...

2

Dear Mr. Cunningham: The question you son asked is at an elementary level, but it has a wide range of associations. I think a one good reference is Prof. Tao's article: https://terrytao.wordpress.com/2020/12/23/246b-notes-1-zeroes-poles-and-factorisation-of-meromorphic-functions/#more-12187 The article should be relatively self-contained. Hope it helps. Some ...

1

Since $F$ is bicomplex-holomorphic, we necessarily have $\frac{\partial F}{\partial Z^\dagger} = 0$, and your condition, let's call it the bicomplex Beltrami equation, $$\frac{\partial F}{\partial Z^\dagger}=\mu(Z)\frac{\partial F}{\partial Z}$$ implies that either $\mu$ is the zero measure or $F$ is constant. By the way, any bicomplex-holomorphic function ...

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