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Search for text files where two different words exist (any order, any line) I'm looking for a way to search files where two word instances exist in the same file.  I've been using the following to perform my searches up to this point:  find . -exec grep -l "FIND ME" {} \;   The problem I'm running into is that if there isn't exactly one space that between "FIND" and "ME", the search result does not yield the file.  How do I adapt the former search string where both words "FIND" and "ME exist in a file as opposed to "FIND ME"?  I'm using AIX. Not getting any reading from FTDI UMFT201XB USB-to-I2C module I got the  in order to interface with an I2C peripheral through a USB port on my Windows 7 computer.  The VCP drivers seem to install correctly, and a COM port is assigned to the module, and can be opened by a terminal emulator. However, I cannot seem to get any signal in or out of the module. That is, I see no activity when I send data through the port, or receive any data when I connect an I2C device to it (this was debugged using a scope). I have also verified that the RESET isn't on.  I found very little helpful information out there on how to get this module to work, so if anyone has experience with it, or with similar FTDI modules, my question is: what are the steps that I need to take in order to get the module to communicate? Does “gay” still include the meaning “merry”? Dictionary.com lists , with “merry, lively” as the first entry.  Microsoft  for listing  (a real place) as his hometown:   Xbox Live considered the term gay absolutely inappropriate in any context.     As a non-native English speaker, my question is whether in contemporary English, does gay always mean homosexual? Are British and American English the same in this respect? Accourding to ,   We often use "been to" when we refer to completed visits to a place:  Have you ever been to Budapest?  Not: Have you ever gone to Budapest?  If the visit is not complete, we use "gone to".  Joan’s just gone to the shop. She’ll be back in about ten minutes. (Joan has not yet returned from the shop.)  Joan’s just been to the shop. She bought some cakes. Would you like one? (Joan has returned from the shop.)  Now, does that rule apply to the  "Have been on a trip" &amp; "Have gone on a trip"?  So, "I have been on a trip" (I completed the trip &amp; currently I am staying at home)  "I have gone on a trip" (I haven't completed the trip &amp; currently I am on the trip)  And what about "gone crazy" &amp; "been crazy".  Does "He has gone crazy" mean now he is still crazy?  Does "He has been crazy" mean now he is not crazy any more?  Are there any differences between "Have been on a trip" vs "Have gone on a trip" and "Have been crazy" vs "Have gone crazy"?  Note: I am not asking about "been to" or "gone to". I am asking about "been on a trip" &amp; "gone on a trip". Do we apply the same rule of "been to" or "gone to" to "been on a trip" &amp; "gone on a trip"?  Note2:  seems to prove that my thinking is right. How to install Windows 7 after Ubuntu and dual boot? If $|G| = pq$,  $p,q$ primes, $p \gt q, q \nmid p-1 $, then how do I prove $G$ is cyclic without using Sylow's theorems? I'm trying to calculate the amount of settings the enigma machine has. I have found several sites regarding this topic, but it seems like there are two answers to my question. The first answer is 158,962,555,217,826,360,000 and the second is 107,458,687,327,250,619,360,000. Where the difference is that in the second answer, it has been taken into account that the notched rings on the enigma could be set as well, giving 26 * 26 = 676 more settings than the first answer.   But which answer is the right one, and why? How many possible different Enigma machine settings are there and how can one determine this number? How to find creation date of file? I want to find out the creation date of particular file, not modification date or access date.  I have tried with ls -ltrh and stat filename. Elementary properties of closure Hi everyone I'd like to know if the following is correct. I really appreciate any suggestion. (Honestly the only one that matters me is the second property the others are easy, I think) Thanks.  Definition: Let $X\subset \mathbb{R}$ and let $x'\in \mathbb{R}$, we say that $x'$ is an adherent point of $X$ iff $\,\forall\varepsilon&gt;0\,\exists x\in X \text{ s.t.} \; d(x',x)\le \varepsilon$. We say that $\overline{X}$ is the closure of $X$ iff contain all the adherent points of $X$.  Lemma (Elementary properties of Closures): Let $X$ and $Y$ be arbitrary subset of $\mathbb{R}$. Then $X\subset \overline{X}, \; \overline{X\cup Y}=\overline{X}\cup\overline{Y},\; \overline{X\cap Y}\subset\overline{X}\cap\overline{Y}$. If $X\subset Y$ then $\overline{X}\subset \overline{Y}$.  Proof:  Clearly if $x\in X$,  $x$ is $\varepsilon$-adherent to $x$ (indeed is $0$-close to itself) for every $\varepsilon$, hence $x\in \overline X$.   Let $z'$ be an adherent point of $X\cup Y$, then there exists some $z\in X\cup Y$ such that $d(z',z)$ for a given $\varepsilon&gt;0$. Suppose  $z\in X$, hence $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}\subset \overline{X}\cup \overline{Y}$. Similarly when  $z\in Y$, $z'\in \overline{Y}\subset \overline{X}\cup \overline{Y}$. Thus $ \overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. We  know that $X\cup Y \subset\overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. Then $\overline{\overline{X\cup Y}}=\overline{X}\cup \overline{Y}$. Since $\overline{\overline{X\cup Y}} = \overline{X\cup Y}$ we're done.  Let $z'\in \overline{X\cap Y}$, i.e., $z'$ is an adherent point of $X\cap Y$. So, there is some $z\in X\cap Y \subset X$ such that $d(z',z)\le \varepsilon$ for any given $\varepsilon&gt;0$. It follows that $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}$. Similarly since $z\in X\cap Y \subset Y$ we have $z'\in \overline{Y}$. Hence $z'\in \overline{X}\cap\overline{Y}$.   Let $x'\in \overline{X}$ we'd like to show that $x'\in \overline{Y}$. Let $\varepsilon&gt;0$ be arbitrary then $\exists x\in X\subset Y$ such that $d(x',x)\le \varepsilon$, so $x'$ is an adherent point of $Y$, i.e., $x'\in \overline{Y}$ as desired. $\Box$     Claim 1: If $X\subset Y \subset \overline{X}$. Then $\overline{Y}=\overline{X}$.    Proof claim: Let $y'$ be an adherent point of $Y$, i.e., $y'\in \overline{Y}$ and let $\varepsilon&gt;0$ be given. Thus, $\exists y \in Y\subset \overline{X}$ s.t. $d(y,y')\le \varepsilon$. So, $y$ is an adherent point of $X$. Thus $\exists x\in X$ s.t. $d(x,y)\le \varepsilon$ and therefore  we must have $d(x,y')\le 2\varepsilon$ which shows that $y'$ is an adherent point of $X$, i.e., $\overline{Y}\subset \overline{X}$. Conversely if $x'$ is an adherent point of $X$. Then $\exists x\in X\subset Y$ s.t. $d(x,x')\le \varepsilon$, and so is adherent to $Y$. $\Box$     Claim 2: $\overline{\overline{X}} = \overline{X}$    Proof: It will suffice to show that if $x''\in \overline{\overline{X}}$, then $x''$ is an adherent point of $X$. Let $\varepsilon &gt;0$ be arbitrary and let $x''$ be an adherent point of $\overline{X}$. Then there is some $x' \in \overline{X}$ such that $d(x'',x')\le \varepsilon$. Since $x' \in \overline{X}$ by definition is an adherent point of $X$, so there is some $x\in X$ for which $d(x,x')\le \varepsilon$. Thus $d(x,x'')\le d(x,x')+d(x',x'')\le 2\varepsilon$ it follows that $x''$ is an adherent point of $X$ as desired. $\Box$ Smaller membership symbol To mean that x is an element of X, we can write x \in X. However, in the standard packages, the membership symbol that results is too big; it draws too much attention to itself. This is especially evident when we're doing set theory and writing sets as lowercase, as in x \in y. Does anyone one how to draw a less attention-seeking membership symbol? I have noticed there is a Stack Overflow in language differs from English which is global and mostly used by programmers.  I think there is no need for a site in other languages. Otherwise we will end up with 127 StackOverflow sites. I am afraid that we will lose the amount of people sharing data. Farm Solution Vs. SandBox Why are there $1 / 1.602176634 \times 10^{-19}$ electrons in a coulomb? Why that exact number of electrons in one coulomb? who decided it? there is nothing wrong with the number, it just seems slightly messy. Why didn't the scientific community just settle on an easier number, such as $1\times10^{-19}$ for example? Flagging accepted "answers" top output: cpu usage > 100% Here is the top output which I gathered:    I noticed that top is showing vlc's cpu usage is > 100%.  Can anybody please justify me why it is showing like this, whether this is a bug in top application or something else. Does Java have buffer overflows? "whatever" as pure determiner? My dictionaries and references define, and I've always thought of, one of the functions of the word whatever as a "relative determiner."  In a sentence like, "I will help you in whatever way is possible,"  inserting the relative pronoun "that" between "way" and  "is", or replacing "whatever" with "the" - a pure, non-relative determiner - (I reckon) renders it ungrammatical:     *I will help you in whatever way that is possible.   *I will help you in the way is possible.   However, I do often encounter utterances like the first one above, such as:     My hope and intent is to contribute in whatever way that I can.   I wish to be of service to the local communities in whatever way that is possible.   where "that" is used, and thus "whatever" can be replaced by "the" without  making the sentences ungrammatical (if they are grammatical to begin with):     My hope and intent is to contribute in the way that I can.   I wish to be of service to the local communities in the way that is possible.   Is this usage of "that" with "whatever" wrong, though colloquial? I want to calculate $$ \int_{0}^{\infty} \frac{\arctan{2t}}{e^{2\pi t}+1} \, {\rm d}t $$ and am wondering if there at all is an analytical expression for the result. I haven't tried much so far, but expressing the denominator as the geometric series and then substituting $z=2\pi t (n+1)$. Equivalently I get $$ \sum_{n=0}^{\infty} \frac{{\rm si}\left(\pi(n+1)\right)}{2\pi (n+1)} $$ where $$ {\rm si}(x) = -\int_{x}^{\infty} \frac{\sin(t)}{t} \, {\rm d}t $$ is the shifted ${\rm Si}(x)$ function. The value is approximately $0.03618247146$, in fact I know from comparison with something else that it is exactly $\frac{\ln \pi -1}{4}$, but there should be some (direct) derivation, or?  Thanks for your input if you have any.  Edit: I tried to follow your instructions, but am now stuck. So I wrote $$ {\cal L}^{-1} \left(\frac{\psi(t)-\psi(1)}{t-1}\right)(s) =\sum_{n=0}^{\infty} \frac{e^{-sn}}{n+1} = e^s \int_s^\infty \frac{{\rm d}x}{e^{x}-1} $$ and taking the Laplace-transform \begin{align} \psi(t) &amp;= \psi(1) + (t-1)\int_0^\infty {\rm d}s \, e^{-s(t-1)} \int_s^\infty \frac{{\rm d}x}{e^{x}-1} \\ &amp;= \psi(1) + (t-1) \int_0^\infty {\rm d}x \int_{0}^{x} {\rm d}s \, \frac{e^{-s(t-1)}}{e^{x}-1} \\ &amp;= \psi(1) + \int_{0}^{\infty} {\rm d}x \, \frac{1-e^{-x(t-1)}}{e^x-1} \tag{1} \\ &amp;= -\frac{1}{t} + \psi(1) + \int_0^\infty {\rm d}x \, \frac{1-e^{-tx}}{e^x-1} \tag{2} \\ &amp;= \dots \\ &amp;=-\sum_{k=0}^{n} \frac{1}{t+k} + \psi(1) + \int_0^\infty {\rm d}x \, \frac{1-e^{-(t+n)x}}{e^x-1} \\ &amp;\stackrel{(\rm{2})}{=} -\frac{1}{2t} + \ln(t) - \int_0^\infty {\rm d}x \, \left\{ \frac{1}{2} - \frac{1}{x} + \frac{1}{e^x-1}\right\} e^{-tx} \tag{Binet 1} \end{align} where $\psi(1)=-\gamma$.  Now we integrate over $t$ from $1$ to $y$ yielding $$ \ln \Gamma(y) \stackrel{({\rm 1})}{=} \psi(1)\,(y-1) + \int_0^\infty {\rm d}x \, \frac{y-1+\frac{e^{-x(y-1)}-1}{x}}{e^x-1} $$ but what now? Substitute? Database does not update automatically with MySQL and Python I'm having some trouble updating a row in a MySQL database. Here is the code I'm trying to run:  import MySQLdb  conn=MySQLdb.connect(host="localhost", user="root", passwd="pass", db="dbname") cursor=conn.cursor()  cursor.execute("UPDATE compinfo SET Co_num=4 WHERE ID=100") cursor.execute("SELECT Co_num FROM compinfo WHERE ID=100") results = cursor.fetchall()  for row in results:     print row[0]  print "Number of rows updated: %d" % cursor.rowcount  cursor.close() conn.close()   The output I get when I run this program is:     4Number of rows updated: 1   It seems like it's working but if I query the database from the MySQL command line interface (CLI) I find that it was not updated at all. However, if from the CLI I enter UPDATE compinfo SET Co_num=4 WHERE ID=100; the database is updated as expected.  What is my problem? I'm running Python 2.5.2 with MySQL 5.1.30 on a Windows box. Why does my Mac's list of "Open with" applications start listing the apps twice and sometimes three times?