there is an exercise in a book on complex analysis:  "Let $f$ is a holomorphic function on $\mathbb{C}$ such that exists $a&gt;0$ with property $f(z) = f(z+a) = f(z+ja)$ for each $z \in \mathbb{C}$. Show, that f must be necessarily constant."  The hint is to use Liouville theorem: "Every bounded entire function must be constant. That is, every holomorphic function f for which there exists a positive number $M$ such that $|f(z)| \le M$ for all $z \in \mathbb{C}$ is constant."  So to finish the exercise it suffices to show $f(z$) is bounded for all $z \in \mathbb{C}$. Could someone help me? Or miss i something trivial?  Thank you.
Let $f: \Bbb C \to \Bbb C$ be an entire (analytic on the whole plane) function such that exists $\omega_1,\omega_2 \in \mathbb{S}^1$, linearly independent over $\Bbb R$ such that: $$f(z+\omega_1)=f(z)=f(z+\omega_2), \quad \forall\,z\in \Bbb C.$$Prove that $f$ is constant.   The intuition seems clear to me, we have the three vertices of a triangle given by $0$, $\omega_1$ and $\omega_2$. All points in the plane are one of the vertices of that triangle under a suitable parallel translation. The constant value will be $f(0)$, fine. Throwing values for $z$ there, I have found that $$f(n\omega_1) = f(\omega_1) = f(0)=f(\omega_2) = f(n\omega_2), \quad \forall\, n \in \Bbb Z.$$  I don't know how to improve the above for, say, rationals (at least). Some another ideas would be:   Checking that $f' \equiv 0$. I don't have a clue of how to do that. Write $w = a\omega_1+b\omega_2$, with $a,b \in \Bbb R$, do stuff and conclude that $f(w) = f(0)$. This approach doesn't seem good, because I only have a weak result with integers above. Finding that $f$ coincides with $f(0)$ on a set with an accumulation point. This seems also bad: the set on with $f$ coincides with $f(0)$ by which I found above is discrete.    Nothing works and this is getting annoying... And I don't see how analyticity comes in there. I'll be very thankful if someone can give me an idea.  (On a side note.. I know that this title is not informative at all. Feel free to edit if you come up with something better.)
If $f:U\to V$ is holomorphic and $f'(z)\neq 0$ for all $z\in U$, then$f$ is locally bijective. I am trying to solve the following problem:     Let $f:U\to V$ be a holomorphic function such that $f'(z)\neq 0$ for all $z\in U$. Show that for all $z_0\in U$, there exists a disc $D_\varepsilon(z_0)\subseteq U$ such that $f:D_\varepsilon(z_0)\to f(D_\varepsilon(z_0))$ is bijective.   Atempt:  I am trying to use Rouche's Theorem, but I am stuck on a particular step. Let $z_0\in U$. Since $f'(z_0)\neq 0$, there is a disc $D_r(z_0)\subseteq U$ such that $$f(z)=f(z_0)+(z-z_0)h(z),\quad\forall z\in D_r(z_0)$$ where $h$ is holomorphic on $D_r(z_0)$ and $h(z)\neq 0$ for all $z\in D_r(z_0)$.   Let $0&lt;\varepsilon&lt;r$ to be defined later and fix $w\in D_\varepsilon(z_0)$. To show that $f$ is injective in $D_\varepsilon(z_0)$ is to show that the function $f(z)-f(w)$ has exactly one zero in $D_\varepsilon(z_0)$. But $$F(z):=(z-z_0)h(z)$$ has exactly one zero in $D_\varepsilon(z_0)$, so we might want to apply Rouche's Theorem with $F$ and $$G(z):=f(z)-f(w)-(z-z_0)h(z)=f(z_0)-f(w)=-(w-z_0)h(w)$$ to conclude that $F$ and $F+G$ have the same number of zeros and hence $f$ is injective. But that means that we need to find $0&lt;\varepsilon&lt;r$ such that $|G(z)|&lt;|F(z)|$ on $\partial D_\varepsilon(z_0)$. That is, $$|w-z_0||h(w)|&lt;|z-z_0||h(z)|$$ for all $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$.  Can we find such $\varepsilon&gt;0$?  It seems intuitive since if we expand $h$ in a power series at $z_0$, $h(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$, then $$|w-z_0||h(w)| = |w-z_0|\sum_{n=0}^\infty a_n|w-z_0|^n &lt; |z-z_0|\sum_{n=0}^\infty a_n|z-z_0|^n.$$ for $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$. But this is of course not a proof. UV Unwrapping a Modified Model I modified an existing model by removing unwanted vertices, then selecting surrounding vertices and creating faces one triangle at a time (By pressing 'F').  On UV unwrapping the mesh, the faces which were added by me seem to stretch from the very bottom left of the map, shown below.    How do I create a face so it appears on the UV map as expected? This dreaded issue started at some time after upgrade to OS X (Yosemite) 10.10.3. accountsd uses about 60% CPU constantly, making the laptop hot and noisy.  I'm not using iCloud at all.  Killing it doesn't help, because it starts again quickly.  Probably related question, but it's about iOS: WARNING DO NOT ATTEMPT TO RUN THIS ON A PRODUCTION MACHINE   In reading the  I generally follow what's going on with the following code:  :(){ :|:&amp; };:   excerpt of description     The following fork bomb was presented as art in 2002;   its exact origin is unknown, but it existed on Usenet prior to 2002.   The bomb is executed by pasting the following 13 characters into a    shell such as  or . It operates by defining   a function called ':', which calls itself twice, once in the   foreground and once in the background.   However the last bit isn't entirely clear to me. I see the function definition:  :(){ ... }   But what else is going on?  Also do other shells such as ksh, csh, and tcsh also suffer the same fate of being able to construct something similar? I'm having trouble understanding the simple voltage regulator that can be built using a zener diode (from section 2.04 in the Art of Electronics).  I know that it would be better to use amplifiers, et cetera, but I'm just trying to understand how this circuit works.      I don't really understand how the circuit works, but I am guessing that when a load is applied to the output, it drains current from the source (Vin) and thus causes the voltage to drop?  How does the zener diode help to maintain the voltage and thus make this circuit act as a regulator? Are "whores" and "horse" homophones? I’m Spanish but sometimes see TV shows in English.  My question is whether the words horse and whores sound exactly the same, because in many English language TV shows it seems like they are, which really surprises me.   Are they homophones?  I cannot hear any voicing at the end of the word whores to distinguish the two words. QGIS corrupting shapefiles? I'm hoping someone out there might be able to help or suggest some tests to try and resolve some very odd QGIS behaviour of late...  Our core GIS team of 4 users has been testing v2.10 prior to roll-out for the remainder of the organization.  In the past couple of weeks we have had some very odd shapefile corruption things happening and are yet to isolate or identify a potential cause...  These have included:   Editing one line (using the scissor tool to split into two records) and having the resulting records attributes completely replaced by attributes from another geometry no-where near the edited row (spatially or in the attribute dbf) Line geometries completely disappear while their attributes get associated with an unrelated geometry  Line geometry get partially deleted and replaced with a vertice in infinity (line visually extends beyond the dataset and no matter how far zoomed out you never see the end)   The above is happening in a polyline layer, that we did discover has some multi-line geometries..? Perhaps this is impacting... but never saw this issue with v2.8 and the same layer. The layer has undergone MANY save-as's in the past week and many verify geometry checks as we thought invalid geometry may have been the cause...  In another separate layer yesterday, to a separate user (me) i created a layer with approx 30 polygons... adding, deleting some and doing some 'avoid intersection' to remove overlaps. Layer looks fine in qgis, saved and also a save-as to a completely new file but when i load into ArcGIS online (and later is ArcMap as a test) some of the overlaps and deleted objects are still there. Close and reopen qgis,  reopen the file and they are definitely not.  The project was created in v2.8 that i was editing in (but had been saved in 2.10).  Does anyone have any ideas/suggestions.  We do still have 2.8 installed as well as 2.10? We have never encountered issues like this when testing other new qgis versions while still running old ones? I know how to rename files in Unix:  $ mv ~/folder/subfolder/file.txt ~/folder/subfolder/file.sh       ^-------this part------^    ^------this part-------^   It takes too long time to repeat ~/folder/subfolder/file twice.  Is there a quicker way? I couldn’t find any proof on the internet. I would appreciate an elementary proof , but any help would be appreciated!  Thank you. How to automaticaly enter sudo's grace period upon login How to pass the output of one command as the command-line argument to another? So I have a script that, when I give it two addresses, will search two HTML links:  echo "http://maps.google.be/maps?saddr\=$1\&amp;daddr\=$2" | sed 's/ /%/g'   I want to send this to wget and then save the output in a file called temp.html. I tried this, but it doesn't work. Can someone explain why and/or give me a solution please?  #!/bin/bash url = echo "http://maps.google.be/maps?saddr\=$1\&amp;daddr\=$2" |  sed 's/ /%/g' wget $url Construction of Dirichlet distribution with Gamma distribution Let $X_1,\dots,X_{k+1}$ be mutually independent random variables, each having a gamma distribution with parameters $\alpha_i,i=1,2,\dots,k+1$ show that $Y_i=\frac{X_i}{X_1+\cdots+X_{k+1}},i=1,\dots,k$, have a joint ditribution as $\text{Dirichlet}(\alpha_1,\alpha_2,\dots,\alpha_k;\alpha_{k+1})$   Joint pdf of $(X_1,\dots,X_{k+1})=\frac{e^{-\sum_{i=1}^{k+1}x_i}x_1^{\alpha_1-1}\dots x_{k+1}^{\alpha_{k+1}-1}}{\Gamma(\alpha_1)\Gamma(\alpha_2)\dots \Gamma(\alpha_{k+1})}$.Then to find joint pdf of $(Y_1,\dots,Y_{k+1})$ I can not find jacobian i.e.$J(\frac{x_1,\dots,x_{k+1}}{y_1,\dots,y_{k+1}})$ To make sure that we are talking about the same, I would like to post the relevant definitions I know first.  Definitions:  A pair $(G, +)$ where $G$ is a set and  $+: G \times G \rightarrow G$  is called a commutative group if it has all of the following three characteristics:   (associativity): $\forall x,y,z \in G: (x+y)+z=x+(y+z)$ (identity element): $\exists e \in G \forall x \in G: e + x = x = x +e$ (inverse elements): $\forall x \in G \exists x^{-1} \in G: x^{-1} + x = e = x + x^{-1}$ (commutativity): $\forall x,y \in G: x+y=y+x$.     A triple $(\mathbb{K}, +, \cdot)$, where $\mathbb{K}$ is a set and  $+: \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$  $\cdot: \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$  is called a field if it has all of the following three characteristics:   $(\mathbb{K}, +)$ is a commutative group $(\mathbb{K} \setminus \{0\}, \cdot)$ is a commutative group distributive properties: $\forall x,y,z \in \mathbb{K}: x \cdot (y+z) = (x \cdot y) + (x \cdot z)$ and $\forall x,y,z \in \mathbb{K}: (y+z) \cdot x = (y \cdot x) + (z \cdot x)$     $\mathbb{N^+} := 1, 2, 3, 4, 5 ...$ (all positive integers)  $\mathbb{N}_0 := \{0\} \cup \mathbb{N}$ (the natural numbers with zero)  $\mathbb{N}$ can be both: $\mathbb{N^+}$ or $\mathbb{N}_0$.    My question: Does a field exist that has $\mathbb{N}$ as its set?  I know that $(\mathbb{Q}, +, \cdot)$ is a field and I know that $(\mathbb{N}, + )$ is only a commutative semigroup.   But maybe it's possible to define two mappings $\circ, *$ for $\mathbb{N}$ in such a way that $(\mathbb{N}, \circ, *)$ is a field.    Related:     The smallest field containing the integers is the field of rational   numbers.   Source: en-Wikipedia:   I don't know if this is true. If somebody can prove (or at least scribble the proof of) the quote it would be a good answer, I guess. Multiple file upload in php titlesec: loss of section numbering with the new update (2016/03/15) After reinstalling TeXLive distribution (install-tl-20160318, Ubuntu 14.04), a issue occurs with the updated version of the titlesec package: the section numbering is lost.  What could it be missing? Is there a modification I have not seen?  Below is a MWE comparison of the two versions. The prevtitlesec package is a reinstall of the previous version (2011/12/15).  \documentclass{book} \usepackage[margin=2cm,a4paper]{geometry} %\usepackage{prevtitlesec}% 2011/12/15 version \usepackage{titlesec}% 2016/03/15 version \usepackage{lipsum}  \begin{document}  \chapter{Title of the first chapter} \lipsum[1] \section{A test section} \lipsum[4] \subsection{A test subsection} \lipsum[4] \subsection{A test subsection} \lipsum[4]  \end{document}   titlesec package 2011/12/15:    titlesec package 2016/03/15: "The antipodal map of $S^2$ is the isometry $\bar m$ sending each point $P$ to its opposite point $-P$" (from textbook)  Show that antipodal points remain antipodal under any isometry of $S^2$. In other words, for each isometry f of $S^2$, if $P$ and $P'$ are antipodal, then so are $f(P)$ and $f(P')$  I am unsure how I can define the $P$ and $P'$ so that they are opposites of each other in the antipodal map, and am I supposed to apply the traditional isometries (translation, reflection and rotation) to the points? I don't understand this problem. What is the difference between CentOS "DVD" vs "Everything" ISOs What does a confidence interval with a negative endpoint mean? Reinhardt's main strength is his shielding abilities and the protection he provides for his team, but Are there any abilities that cannot be blocked by Reinhardt's barrier field? Creating Workflow -> Error