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Extracting the function from InterpolatingFunction object Let $A$ be a commutative ring and $S$ a multiplicative closed subset of $A$. If $A$ is a PID, show that $S^{-1}A$ is a PID.   I've taken an ideal $I$ of $S^{-1}A$ and I've tried to see that is generated by one element; the ideal $I$ has the form $S^{-1}J$ with $J$ an ideal of $A$. $J$ is generated by one element but I can't see why $I$ has to be generated by one element, maybe I'm wrong. Spacing before and after section titles I know this subject has been mentioned before, but I'm still having a problem in creating a space under a section title. I used the titlesec package and then tried using the command \titlespacing{\section}...  This is my code. Can anybody help?  \documentclass[10pt]{book} \usepackage[utf8]{inputenc} \usepackage{fullpage} \usepackage[compact]{titlesec} \usepackage{setspace} \usepackage{sectsty} \chapterfont{\huge} \sectionfont{\Huge} \begin{document} \chapter*{Kapitel 1} \section*{Einleitung} \begin{large} \subsection*{1.1 Motivation} I am drawing a 3D array like the attached picture, which actually describes a multidimensional array. However, I can not put the name of the axis and take the value from the matrix like the attached picture.  This is the code that I have:  \documentclass[margin=5mm]{standalone} \usepackage{tikz} \usetikzlibrary{matrix,calc}  \begin{document} \begin{tikzpicture}[every node/.style={anchor=north east,fill=white,minimum width=1.4cm,minimum height=7mm}] \matrix (mA) [draw,matrix of math nodes] { (1,1,3) &amp; (1,1,3) &amp; (1,1,3) &amp; (1,1,3) \\ (1,1,3) &amp; (1,1,3) &amp; (1,1,3) &amp; (1,1,3) \\ (1,1,3) &amp; (1,1,3) &amp; (1,1,3) &amp; (1,1,3) \\ (1,1,3) &amp; (1,1,3) &amp; (1,1,3) &amp; (1,1,3) \\ };  \matrix (mB) [draw,matrix of math nodes] at ($(mA.south west)+(1.5,0.7)$) { (1,1,3) &amp; (1,1,3) &amp; (1,1,3) &amp; (1,1,3) \\ (1,1,3) &amp; (1,1,3) &amp; (1,1,3) &amp; (1,1,3) \\ (1,1,3) &amp; (1,1,3) &amp; (1,1,3) &amp; (1,1,3) \\ (1,1,3) &amp; (1,1,3) &amp; (1,1,3) &amp; (1,1,3) \\ };  \matrix (mC) [draw,matrix of math nodes] at ($(mB.south west)+(1.5,0.7)$) { (1,1,3) &amp; (1,1,3) &amp; (1,1,3) &amp; (1,1,3) \\ (1,1,3) &amp; (1,1,3) &amp; (1,1,3) &amp; (1,1,3) \\ (1,1,3) &amp; (1,1,3) &amp; (1,1,3) &amp; (1,1,3) \\ (1,1,3) &amp; (1,1,3) &amp; (1,1,3) &amp; (1,1,3) \\ };  \draw[dashed](mA.north east)--(mC.north east); \draw[dashed](mA.north west)--(mC.north west); \draw[dashed](mA.south east)--(mC.south east); \end{tikzpicture}   \end{document} Modify iPhone restrictions passcode Is there a way to change the restrictions passcode on an iPhone without restoring the phone? I haven't forgotten the passocde; I just want to change it. Let $A$ be a closed and bounded subset of $\mathbb{R}^{n}$. Given that $f: A \to A$ satisfies $d(f(x),f(y)) &lt; d(x,y)$ for all $x,y \in A$, $x\neq y$ where $d$ is the usual Euclidean metric. Show that $f$ has a unique fixed point.    My observations:    Since $\mathbb{R^{n}}$ is complete, then $(A,d)$ is complete since $A$ is closed; Since $A$ is bounded, then $A \subset B_{R}$ for some $R &gt; 0$; Banach Fixed Point Theorem guarantees a unique fixed point if $f$ is contraction on $A$.   Please correct me if I am wrong. My quandry here is showing that $f$ is a contraction, i.e.  $$ d(f(x),f(y)) \leqslant k\cdot d(x,y), \hspace{3mm} k \in (0,1) $$  I'm looking for some hints or tips. Thank you. USB-C Dongles killing Wifi on 2016/2017 MacBooks – Who’s to blame? I recently switched from my beloved MacBook Pro Retina (Late 2013) i5 2.6/16/512 to MacBook Pro (2017) Non-TB i5 2.3/16/512.  I purchased what from the build-quality looked like a good USB-C-to-DisplayPort cable with 4k/60Hz support, just to run into the very problem the Logicboard repair guru Louis Rossmann had described in : As soon as you plug in the USB-C dongle, the (2.4 GHz) Wifi stops working despite full antenna signal. Unplug the USB-C connector and you are immediately back in the game.  The big question I'm asking myself now: Who is to blame? Is this a design flaw in the current generation MacBooks with improper shielding or is it a design flaw in the various dongles?  I mean the fact that even Apple’s very own USB-C Digital AV Multiport Adapter is causing these WiFi issues (as you can learn from the many horrific ; but AFAIK have been fixed through a ), show that there’s something going horribly wrong. I have white blank on the posts.    App Version: 1.0.89 Device Manufacturer: Meizu Device Model: m2   OS Version: 5.1 (1464689330)    I can't see the questions and the answers and the comments.  I saw that it's a common issue..  Does it have a fix?   EDIT -  I saw  answer. This answer is ~2 years old. It says that it will fix in the next update.  His version is 1.0.59 - mine is 1.0.89. Huge different.   The problem is still exists! Can I skip over releases when upgrading? I recall having heard somewhere that the only 1-manifolds (second countable, Hausdorff, connected spaces locally homeomorphic to $\mathbb R$) are $\mathbb R$ and $S^1$. Is this true? If so, is there a reasonably elementary proof? When I ordered this TP-Link AC1300 Wireless Dual Band PCI-Express Adapter (Archer T6E), I looked into customers reviews and they said it worked fine with no drivers on Ubuntu. I didn't have a CD/DVD player on my desktop so I (on a laptop with a DVD drive) copied and pasted the DVD support disk content on to a USB drive and used that. But when I tried to run autorun.exe it puts up archive manager and says An error occurred while loading the archive Ubuntu verison- Ubuntu 16.04 LTS.  Output from: lspci -vnn  02:00.0 Ethernet controller [0200]: Realtek Semiconductor Co., Ltd. RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller [10ec:8168] (rev 11) Subsystem: ASRock Incorporation Motherboard (one of many) [1849:8168] Kernel driver in use: r8169  Kernel modules: r8169   I am fairly new to Ubuntu (1 month) and have learned alot (sudo commands) and now know that windows is overpriced and that ubuntu is better in most fields.  Thanks for the help. Contraction Map on Compact Normed Space has a Fixed Point Custom Login form question In my website have two types people login one is designer another is reader, i want people come to login my website by role. Mean designer go to design login form to login. Reader go to reader login form to login. my question is:   how to show two login forms in same page? how to check the role when the visiter login at the same time in login form? is it any module i can use or i need do something customize the user module and how    customize?   Plz help! How do you select a particular option in a SELECT element in jQuery? Wi-Fi not working on Lenovo ThinkPad E570 (Realtek RTL8821CE) We just bought a Lenovo  (which is supposed to have the Intel 8265 for wireless). This laptop was bought in part because it is certified to run Ubuntu and as such to me at least it would make sense for things to work out of the box... but no such luck.  First, I installed Ubuntu 17.10 and after installation realised Wi-Fi was not working. I also tried Ubuntu 16.04 from a USB drive since that is the version of Ubuntu it is certified for, but no luck there either.  The wireless chip is not detected by lshw (or even lspci by the looks of it), ip link does not show a wireless interface, and the iwlwifi kernel module is not loaded at boot (I can load this myself using modprobe iwlwifi but this does not make Wi-Fi work). I think it is a combined Wi-Fi/Bluetooth chip, and Bluetooth does appear to work (or at least Bluetooth shows up in Gnome and rfkill list).  modinfo iwlwifi | grep 8265 shows me a file ending in -34 is supposedly loaded but only a file ending in -33 (and some other lower numbers) is present in /lib/firmware, if that helps).  I have also installed the Linux 4.14.9 kernel in an attempt to get things to work, but this did not resolve the issue either.  Further information  root@ThinkPad-E570:~# modinfo iwlwifi | grep 8265 firmware:       iwlwifi-8265-34.ucode root@ThinkPad-E570:~# rfkill list all 0: tpacpi_bluetooth_sw: Bluetooth     Soft blocked: no     Hard blocked: no 1: hci0: Bluetooth     Soft blocked: no     Hard blocked: no root@ThinkPad-E570:~# dmesg | grep iwl root@ThinkPad-E570:~# lspci -nnk | grep 0280 -A3 05:00.0 Network controller [0280]: Realtek Semiconductor Co., Ltd. Device [10ec:c821]     Subsystem: Lenovo Device [17aa:c024]   Update  I just noticed that the laptop does not have an Intel wireless chip at all, but in fact has a Realtek one... which means that the  of the shop I bought this laptop at were not accurate... Sigh. Same problem stands, however, just with a crappier wireless chip. At what moment will matter falling into a black hole affect its size? Does the mass of matter falling into a black hole affect the size of an event horizon the moment it passes through it, or when it has been incorporated into the singularity? I have to prove that this is a metric: $$d(x,y):=\sum_{i=0}^\infty \frac{|x_i -y_i|}{2^i (1+|x_i-y_i|)}$$  The only thing that I can't prove is the triangle inequality, it's really hard. I know that $|x_i -y_i|\le |x_i -z_i|+|z_i -y_i|$, and I want to get: $$\frac{1}{2^i (1+|x_i-y_i|)} \le \frac{1}{2^i (1+|x_i-z_i|)}+\frac{1}{2^i (1+|z_i-y_i|)}$$ (I think...) The closest thing I've got is: $2^i (1+|x_i-y_i|)\le 2^i (1+|x_i-z_i|)+2^i (1+|z_i-y_i|)$ but... that's not quite what i want. I don't know how to get this... Consider an irreducible but possibly periodic Markov chain on a finite state space with transition matrix $P$. We know there exists a unique stationary distribution $\pi$. If the Markov chain were aperiodic, we would have $P^n_{ij} \to \pi(j)$ as $n \to \infty$. This fails if the chain is periodic, but we have convergence of the Cesaro averages: $$\frac{1}{n}\sum_{k=1}^n P^k_{ij} \to \pi(j) \text{ as } n \to \infty.$$ Can anyone point me to a reference that states this fact? Every reference I've seen only considers convergence for aperiodic chains, or "fixes" periodicity by considering a lazy version of the chain. Alternatively, is there a simple way to obtain this result using the result for aperiodic chains? Show that $\ker \hat{T} = \text{ann}(\text{range } T)$ This is an old exam problem: Let $V$ and $W$ be finite dimensional vector spaces over a field $F$ and let $T: V \to W$ be a linear transformation. Define $\hat{T}: W^* \to V^*$ by $(\hat{T}(f))(v)=f(T(v))$. Here, $U^*$ denotes the dual vector space $U$. Since $\hat{T}$ is a linear transformation prove $\ker \hat{T} = \text{ann}(\text{range } T)$ where $\text{ann}(Y) = \{g \in U^* : g(y)=0 \text{ for all } y \in Y \}$ for a subspace $Y$ of vector space $U$.  I'm really not sure how to start this problem. Any advice would be great! Write to FIFO only if it exists