I have a view listing nodes. I added 2 exposed filters. One is Content: title and the other is Content:Body. Then I have 2 input fields to enter some keyword to do searchs.   I need to have just one field and do the search in both fields: title and body. Can I do this?
I have two user profile fields "Last Name" and "First Name". I also have a user list views. I want to expose "Name" as filter which can search for both Last Name and First Name. How can I do single filter of combination of those two fields? Can I create it from views UI?  One possible way is to create another profile field "Name" which will be hidden in the form. On user save, I would combine the two field values into the "Name" field and then expose it as filter in the views. But this solution is hard coding and it needs to write hooks.
Categorical Principal Component Analysis - using Count, Continuous, Ordinal variables together I have some variables and I want to reduce their number for further analysis. I initially thought of combining them using factor analysis. But since the variables are of all kinds (rating, count, ordinal, continuous dollar amount) I am thinking about using CATPCA for it. However, I have some questions about the technique -   Is CATPCA the right approach or Latent Class Factor Analysis would be the better one? By using the different kinds of variables together - how will the object scores be interpreted? For example - The rating question is on a 10 point scale but the dollar amount ranges from 0 to 1000s, so what will the object score obtained after CATPCA represent? And is it comparable to the Factor Score obtained from traditional PCA? For ordinal variables - Do all of them need to be in the same form, e.g. 1 meaning lowest and 10 meaning highest? Or can they be used in whatever form they are? Is it good to categorize the count and continuous variables based on their distribution and then treat then as ordinal/categorical variables?   I would highly appreciate any help you can provide! Special Relativity - Regarding the Simultaneity of Events During the Train Paradox My questions concerns that classic train paradox, wherein there is a train and a tunnel of equal length, and the train is traveling and some fraction of the speed of light towards the tunnel.  According to the Special Theory of Relativity, an observer outside the tunnel will see the train length contracted (Lorentz Contraction), whereas an observer inside will see the tunnel contracted.   Additionally, suppose that there were doors at the ends of the tunnel and that the observer outside the tunnel closed both doors instantaneously when he/she saw that the train was completely inside the tunnel.  The classic resolution of this paradox invokes the non-simultaneity of events, explaining that the observer in the train sees the far door close first, and then, once the train has begun to exit the tunnel, looks back to see the door at the beginning close. Thus, both observers agree that the train does not touch the door when they are closed for an instant.  Now my questions.  Why is it that the observer on the train sees the far door close first? It seems to me that the information coming from the far door would reach the observer on the train only after the information from the other door is reach.   Under this interpretation, the observer on the train would observe the train getting hit by the doors. What if, by some means, this event can be explained in terms of a stationary observer too? Everyone always concludes that the train remains untouched by the doors, but really the only condition that needs to be met is that both observers must agree. Why can't they both agree that the train was hit?  So, to summarize.   Why is the door that is farther away from the train observed to close first? Why can't the other possible conclusion (both see a hit train) be observed? Prove that $\frac4{abcd} \geq \frac a b + \frac bc + \frac cd +\frac d a$ Let $a, b, c$ and $d$ be positive real numbers such that $a+b+c+d = 4$. Prove that $$\frac4{abcd} \geq \frac a b + \frac bc + \frac cd +\frac d a .$$   How can I approach this using only the ? Are there any other methods that do not involve the AM-GM inequality? Install Numpy for Python3.4.0 on Ubuntu 14.04.01 LTS I'm currently unable to access anything via the Stack Exchange app under iOS; is this a known issue? How do I deal with 'sudoers.so must be only be writable by owner'? Whenever I try to install or remove something using terminal it gives the following error message:  bluebird@bluebird:~$ sudo apt-get update sudo: /usr/lib/sudo/sudoers.so must be only be writable by owner sudo: fatal error, unable to load plugins bluebird@bluebird:~$ sudo apt-get remove vlc sudo: /usr/lib/sudo/sudoers.so must be only be writable by owner sudo: fatal error, unable to load plugins bluebird@bluebird:~$ sudo apt-get install apt-file sudo: /usr/lib/sudo/sudoers.so must be only be writable by owner sudo: fatal error, unable to load plugins I am having difficulty with this proof. I have previously proved that for an automorphism on the extension field $K$ of $F$ that when restricted to $F$ is the identity automorphism must take roots of $f(x)\in F[x]$ in $E$ to roots of $f(x)$ in $E$. I was thinking of using the minimal polynomial $x^{3}-2\in\mathbb{Q}[x]$ and the result of previous problem I solved, but I suspect there are automorphisms that don't fix the subfield, so the result can't be used or I need to prove that any automorphism of an extension field must fix the subfield that is being extended or always sends roots in $E$ to roots in $E$.  I am not asking for a solution to this, but maybe a hint or pointer in the right direction. This problem is introduced in the text prior to Galois theory, so please keep that in consideration. Edit record from site.com In site.com I have a page which iterates records of an object with data repeater.  I want to edit these records from this page ,but I didn't find anything to achieve this.  If there is any option to do this please help me to implement this.  Thanks I changed permissions of a file (chmod g+w testfile) and running ls -l testfile gives:  -rwxrwxr-x 1 user1 user1 0 2011-01-24 20:36 testfile   I then added a user to that group ("/etc/group" has user1:x:1000:user2 line), but am failing to edit that file as user2. Why is this so? Disallow downvoting other answers if I've posted my own Possible Duplicates:        As Dan Dyer mentioned in his comment on , there is a conflict of interest when you have posted your own answer and are allowed to downvote other answers.  Despite the vote editing window being drastically reduced, the fundamental problem remains: I can downvote the other answers permanently and still benefit - my answer gets upvoted ONCE and wham, I have just made up the rep loss for my 10 other downvotes. The vote editing window fixes part of the problem, but there is a more complete fix that has the extra benefit of not preventing legitimate vote changes: to disallow downvotes on competing answers when you have posted one yourself.  The temptation to wrongfully downvote competing answers is huge. The rep loss is tiny in comparison. It has to be in order for people to continue to legitimately downvote.  Details:   If I have previously voted and then post my own answers, delete my downvotes. Continue to allow me to remove the upvotes - I think this is fully legitimate. If I delete my answer it's probably best to continue to deny downvotes (perhaps even including the question), but I don't have a strong opinion either way. Proof of $\lim_{n\to \infty} \sqrt[n]{n}=1$ Thomson et al. provide a proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$ in . It has to do with using an inequality that relies on the binomial theorem:  I have an alternative proof that I know (from elsewhere) as follows.  Proof. \begin{align} \lim_{n\rightarrow \infty} \frac{ \log n}{n} = 0 \end{align} Then using this, I can instead prove: \begin{align} \lim_{n\rightarrow \infty} \sqrt[n]{n} &amp;= \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \newline &amp; = \exp{0} \newline &amp; = 1 \end{align}  On the one hand, it seems like a valid proof to me. On the other hand, I know I should be careful with infinite sequences. The step I'm most unsure of is: \begin{align} \lim_{n\rightarrow \infty} \sqrt[n]{n} = \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \end{align} I know such an identity would hold for bounded $n$ but I'm not sure I can use this identity when $n\rightarrow \infty$. Question: If I am correct, then would there be any cases where I would be wrong? Specifically, given any sequence $x_n$, can I always assume: \begin{align} \lim_{n\rightarrow \infty} x_n = \lim_{n\rightarrow \infty} \exp(\log x_n) \end{align} Or are there sequences that invalidate that identity?  (Edited to expand the last question) given any sequence $x_n$, can I always assume: \begin{align} \lim_{n\rightarrow \infty} x_n &amp;=  \exp(\log \lim_{n\rightarrow \infty} x_n) \newline &amp;=  \exp(\lim_{n\rightarrow \infty} \log x_n) \newline &amp;=  \lim_{n\rightarrow \infty} \exp( \log x_n) \end{align} Or are there sequences that invalidate any of the above identities? (Edited to repurpose this question). Please also feel free to add different proofs of $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$. I'm very familiar with how symbols display differently in inline mode (with $ ... $) vs. how display math shows them (with \[ ... \] or $$ ... $$).  Two examples would be with limits  $\lim_{n\rightarrow \infty}f(x)$   vs.  \[ \lim_{n\rightarrow \infty}f(x) \]   and with sums  $\sum_{n=1}^{x} n^2$   vs.  \[ \sum_{n=1}^{x} n^2 \]   My Question is:     How do you display these things in inline mode (with all the nice inline formatting) as if they were in display mode ?   The inverse -- how to display things in display mode as if they were in inline mode -- is also of interest. Shower at Heathrow during layover in Terminal 5 Reading the  questions here, I am not too hopeful, but this is the perfect site to find the answer if it does exist:  Facts:   I arrive at T5 (British Airways) at 08:30, depart at 11:40 from T5 (British Airways) to another country. No time once I reach final destination to shower, Leave straight from work to catch the first leg. US citizen + passport, only final destination visa (currently)   Desire:  I want to shower - 10 mins is plenty.  Assume money is no object (pay for lounge access, cabs etc), but I'd rather not buy full ticket if I can avoid it.  P.S. It's a wedding, so I REALLY want to shower before haha.  EDIT: My plan so far (which doesn't work because it's after Security):   Travel from  Use    UPDATE:  I followed G &amp; Doc's advice and walked over to the Sofitel Hotel (Immigration + walk ~20-25 min total) showered there for £25 and walked back through Security with plenty of time to spare.  Thank you all for your help! Is every invertible matrix a change of basis matrix? I have noticed that while asking questions and scrolling to the bottom of the page, the right sidebar containing recent questions covers the footer partially. Also, a part of it goes beyond the browser's boundary. It may not be a big problem but it still doesn't look good.    This also happens on the  page I have a process I can't kill with kill -9 &lt;pid&gt;. What's the problem in such a case, especially since I am the owner of that process. I thought nothing could evade that kill option. What does the 'cosmological constant' represent? Newtons theory of gravity involves a gravitational constant $G$, however one does not refer to it directly, we speak instead of gravity or the force of gravity.   Now Einsteins introduced a  in his equation for GR, and ever since it's been referred to as such; however, this usage seems strange to me, given the above parallel; it seems to me we ought to be referring to something physical, which can then be quantified and whose quantification involves the cosmological constant.   Does the cosmological constant then refer to a cosmological force? A force that is, contra gravity, repulsive? And which is even weaker than gravity, kicking in on the scale of galaxies rather than solar systems? Combinatorial argument to prove the recurrence relation for number of derangements Give a combinatorial argument to prove that the number of derangements satisfies the following relation: $$d_n = (n − 1)(d_{n−1} + d_{n−2})$$ for $n \geq 2$.  I am able to prove this algebraically but not able to see the combinatorial example. How many generator has a cyclic group of order n? If you give a user sudo privileges how can I remove the sudo privileges and make the user become just a regular user?  I used:  sudo adduser username sudo   Now I've changed my mind.