If packagehello does not match, the output is still displayed.  Aim: to see no output in situation 2  Situation 1:  user@hostname ~]$ sudo yum list 'package*' packagehello packagehello package2world packagehello package2world   Situation 2:  user@hostname ~]$ sudo yum list 'package*' | grep -E 'package1.*|package2.*' package2world package2world   How to show the output only if both words match using grep?
I would like to get the multi pattern match with implicit AND between patterns, i.e. equivalent to running several greps in a sequence:  grep pattern1 | grep pattern2 | ...   So how to convert it to something like?  grep pattern1 &amp; pattern2 &amp; pattern3   I would like to use single grep because I am building arguments dynamically, so everything has to fit in one string. Using filter is system feature, not grep, so it is not an argument for it.    Don't confuse this question with:  grep "pattern1\|pattern2\|..."   This is an OR multi pattern match.
How do I address unchecked cast warnings? Here's my preamble:  \documentclass[11pt]{amsart} \usepackage{geometry}                % See geometry.pdf to learn the layout options. There are lots. \geometry{letterpaper}                   % ... or a4paper or a5paper or ...  %\geometry{landscape}                % Activate for for rotated page geometry %\usepackage[parfill]{parskip}    % Activate to begin paragraphs with an empty line rather than an indent \usepackage{graphicx} \usepackage{amssymb} \usepackage{epstopdf} \usepackage{float} \usepackage[normalem]{ulem}  \DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}  \title{\emph{Research Proposal:}\\Perceptual Effects of Cross-Contextualized Soundscapes within the Built Environment}   Here's what happens:    Any ideas? Characterizing all ring homomorphisms $C[0,1]\to\mathbb{R}$. This is something I've been trying to work out this evening.      Let $R$ be the ring of continuous real-valued functions on $[0,1]$ with pointwise addition and multiplication. For $t\in [0,1]$, the map $\phi_t\colon f\to f(t)$ is a ring homomorphism of $R$ to $\mathbb{R}$. I'm trying to show that every ring homomorphism of $R\to\mathbb{R}$ has this form.   Suppose otherwise, that there is some $\phi\neq\phi_t$, and thus there is some $f_t\in R$ such that $\phi(f_t)\neq \phi_t(f_t)=f_t(t)$. Define $g_t=f_t-\phi(f_t)1\in R$. Here $\phi(f_t)1$ is the constant function sending $[0,1]$ to $\phi(f_t)$. Then $g_t(t)\neq 0$. My first small question is why does $\phi(g_t)=0$? It seems only that $\phi(g_t)=\phi(f_t)-\phi(\phi(f_t)1)$.  I would like to conclude that there are only finitely many $t_i$ such that $g(x)=\sum g_{t_i}^2(x)\neq 0$ for all $x$. Then $g^{-1}=1/g(x)\in R$, but $\phi(g)=0$, contradicting the fact that homomorphisms map units to units. How can we be sure there are only finitely many $g_{t_i}$ such that the sum of their squares is never $0$? Thanks. Unstable wifi with intel 8260 [8086:24f3] (rev 3a) on Ubuntu 16.04 running on Lenovo P50 Database Encryption scan for database  was aborted. Reissue ALTER DB to resume the scan How to create a torus with divided cuts that correspond to the direction of the torus Navier-Stokes system ODBC connection to Access database in 64-bit Windows Function whose image of every open interval is $(-\infty,\infty)$ How to find a function from reals to reals such that the image of every open interval is the whole of R?  Is there one which maps rationals to rationals? Solvability of Poisson's equation for $p$-forms directly from Hodge decomposition If $M$ is a closed manifold, it's well known that $\Delta\phi=\alpha$, where $\alpha$ is a $p$-form, has a solution if and only if $\alpha$ is orthogonal to the space of harmonic $p$-forms. This is proved on page 224 of Warner, but the proof depends heavily on his approach to Hodge theory. There are other methods that avoid his main technical lemma 6.6. So, can one prove this fact using only the "standard" facts of the theory?  For functions it is easy, for one has $$\alpha=\delta \psi+h.$$ But $(\alpha,h)=0$, so $\alpha=\delta\psi$. Then decompose $\psi$ to get $\alpha=\delta d\zeta$ for some function $\zeta$. But on functions, $\Delta=\delta d$. But for $p$-forms in general one has $$\alpha=d\xi+\delta \psi+h,$$ since $\Omega^{p-1}$ is nontrivial. Using the decompositions again, one has $$\alpha=d\delta \rho+\delta d\sigma,$$ where $\rho,\sigma$ are $p$-forms. I don't see how $\rho$ and $\sigma$ are related, but it looks close. In some literature, I have read that a regression with multiple explanatory variables, if in different units, needed to be standardized.  (Standardizing consists in subtracting the mean and dividing by the standard deviation.)  In which other cases do I need to standardize my data?  Are there cases in which I should only center my data (i.e., without dividing by standard deviation)? Empty intersection and empty union If $A_\alpha$ are subsets of a set $S$ then   $\bigcup_{\alpha \in I}A_\alpha$ = "all $x \in S$ so that $x$ is in at least one $A_\alpha$"  $\bigcap_{\alpha \in I} A_\alpha$ = "all $x \in S$ so that $x$ is in all $A_\alpha$"  It is the convention that $\bigcup_{\alpha \in \emptyset}A_\alpha = \emptyset$ and $\bigcap_{\alpha \in \emptyset} A_\alpha = S$.   But if $x$ is in $\bigcap_{\alpha \in \emptyset} A_\alpha = S$ then $x$ is in all $A_\alpha$ with $\alpha \in \emptyset$ and therefore $x$ is certainly in at least one $A_\alpha$ with $\alpha  \in \emptyset$. But then $x \in \bigcup_{\alpha \in I}A_\alpha$.  Can someone help me and tell me what is wrong with this? Thank you. Kolmogorov proved, that, as one considers independent (not necessary equally distributed) Random Variables: $\{X_n\}_{n\ge0}\subseteq \mathcal L^2$   With $\mathrm{Var} (X_n)=\sigma^2_n$ and without loss of generality $E[X_n]=0$.  If $\sum_{n=0}^\infty \frac{\sigma^2_n}{n^2} \lt \infty$ then SLLN holds, that is:  $$\frac1n\sum_{k=0}^n X_k \rightarrow 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\text{a.s.}$$  Something different:  Also it is known, that if $\sup_{n\ge0}\sigma^2_n =: v \lt \infty$ pairwise uncorrelated $(X_n)_{n\ge0}$ will suffice for SLLN to hold.  My question in view of this situation: Is it possible to weaken the condition of $(X_n)_{n\ge0}$ beeing independent and instead merely consider pairwise uncorrelated $(X_n)_{n\ge0}$? Symfony plugin sfDoctrineActAsTaggablePlugin not working I'm trying to solve a question where I need to write $\frac {1}{1+z^2} $ as a power series centered at $z_0=1$  I'm not allowed to use taylor expansion. So my first thought was to rewrite the function in a form where I can apply the basic identity:  $$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n $$  So let's rewrite the original function $f(z) = \frac {1}{1+z^2} $ and use above identity:  $$\frac {1}{1+z^2} = \frac {1}{1-(-z^2)} $$ But observe that I need to center it at $z_0=1$ so I rewrote it to:  $$\frac {1}{1-(-z^2)}=\frac {1}{2z-(-(z-1)^2)} $$  I can factor out $\frac {1}{2z}$ and use the basic identity to get : $$\frac {1}{2z}\frac {1}{1-\frac{(z-1)^2}{2z}} = \frac {1}{2z}\sum_{n=0}^\infty \left(\frac{-(z-1)^2}{2z}\right)^n =\frac {1}{2z}\sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^n} = \sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^{n+1}}$$  The problem is if I ask wolfram alpha: power series$ \frac{1}{1+z^2} $centered at z = 1 it will give me a different answer that doesn't match the sum I found.  Can anyone help me figure this one out? What am I doing wrong? Was mathematics invented or discovered? What would it mean to say that mathematics was invented and how would this be different from saying mathematics was discovered?   Is this even a serious philosophical question or just a meaningless/tautological linguistic ambiguity? Add Appendix with correct page to ToC Incredible frequency of careless mistakes What is a concise function that formats a (String) decimal into a currency format in Apex? goal: write the most concise function that formats a (String) decimal into a currency format in Apex. Feel free to trainwreck the code for points, I've kept mine exploded for readability purposes. I have 16GB RAM. Do I need 32GB swap?