When I click "log out" the site shows me as being logged out, but I can log back in again by just choosing "Google" option from the log in page. I'm not taken to google.com to enter email/password again.    I've tested to repro, as below.     Before starting each test, I started with a clean browser - no cache, cookies, history - I wiped everything "log out" was done by the drop down menu OS: Debian 8 Browser: Iceweasel 31.8.0     TEST 1   Clicked "log in" on MSE Redirected to MSE login page (3 options) Clicked "Google" Redirected to google.com (login form) Entered credentials, was redirected back to MSE (was logged in) Clicked "log out" on MSE Redirected to log out confirmation page (clicked "log out") Redirected back to main MSE page shown as being logged out Click "log in" Clicked "Google" on login page Am logged in     The problem  After logging out and the site showing me I am no longer logged in, then clicking "log in" I am not redirected to Google to enter my credentials (email/password), when I should be as I had previously clicked "log out" on Stack site.    This means if I click "log out", someone/anyone with access to my computer can click "log in" and then click "Google" and are logged in to my account. I don't have this security issue personally, but many users might with shared devices, public computers, shared work PCs, etc.    Regardless of potential users this affects, I would have thought clicking "log out" would mean if I click "log in" on the same browser I would have to enter my credentials again with Google.      Other tests  TEST 2  The same happens on Stack Overflow following "TEST 1" exactly (so this not MSE specific).    TEST 3  It doesn't matter which site I log in and out of at all. Once I've been logged in, after clicking "log out", I can click "log in" and choose "Google" and without having to enter my credentials I am logged in to all sites again.    Even if I log out on one site, go to a site I've not been to yet, and click "log in" there.    Other  I don't have a Facebook or Stack account, only log in through Google, so cannot confirm if this is specific to Google auth or not.      Cookies  (This is mostly info Stack will use if needed/if it helps).    Based on TEST 1.    Before starting TEST 1:    0 cookies.    After logging in to MSE first time:    All cookies have the same following info:  Host: .meta.stackexchange.com Path: / Expires: At end of session Secure: No   Individual cookies (all contained the above info, but removed from below to shorten list; also removed "value" as it's pointless info):    Name: __cfduid HttpOnly: Yes  Name: __qca HttpOnly: No  Name: _ga HttpOnly: No  Name: _gat HttpOnly: No  Name: acct HttpOnly: Yes  Name: prov HttpOnly: Yes   After clicking "log out":    The same as above except cookie "acct" was no longer there, and an additional one only appeared after I'd clicked "log out":    Name: gauthed Value: 1 Host: meta.stackexchange.com Path: / Expires At end of session Secure: No HttpOnly: No   When I log in again (having logged out but now without having to re-enter credentials to log in):    "gauthed" is gone and "acct" returns.    "acct" is a different value to what it was when I was initially logged in.    The cookie "values" when I was first logged in are the same as when I have clicked log out (except for "acct" as that has gone) and the same as when I log in the second time (except "acct" which changes).
Why does logging in to Stack Overflow keep me signed in to my Gmail email after logging out and regardless of the Stay signed in checkbox?  Steps to reproduce:   Browse to  and confirm you are logged out Browse to  and confirm you are logged out Click login Click Google Enter Email/Password and confirm Stay signed in is unchecked Click Sign in Click logout (up top) Click Logout (down below) Browse to  and confirm you are still logged in Click Sign out &lt;- Shouldn't clicking Logout on Stack Overflow be doing the same thing as this? Browse to  and confirm you are logged out     Answers are making sense but my frustration is that clicking Logout (twice) on StackOverflow essentially does nothing. If I proceed to click login and Gmail I am immediately authenticated because I still have an active Gmail cookie. In order to effectively logout I need to click Logout twice on Stack Overflow and then manually browse to  and Sign out.
A bob hanging in an accelerating train moves backward. What is the force moving it backward? When the train accelerates, the bob would also accelerate with the same magnitude and direction as the train. From the free body diagram, only tension and weight are exerted on the bob. I understand how to relate the "horizontal component" of the tension to the acceleration of the bob. It moves "backward" and in fact its vertical position is different from when the train is not accelerating.  I am not sure whether inertia could explain this situation. What is the force moving it backward/ lifting with the height? Why do Google search results include pages disallowed in robots.txt? I have some pages on my site that I want to keep search engines away from, so I disallowed them in my robots.txt file like this: User-Agent: * Disallow: /email  Yet I recently noticed that Google still sometimes returns links to those pages in their search results.  Why does this happen, and how can I stop it? Background: Several years ago, I made a simple web site for a club a relative of mine was involved in.  They wanted to have e-mail links on their pages, so, to try and keep those e-mail addresses from ending up on too many spam lists, instead of using direct mailto: links I made those links point to a simple redirector /  script running on my own site.  This script would return either a 301 redirect to the actual mailto: URL, or, if it detected a suspicious access pattern, a page containing lots of random fake e-mail addresses and links to more such pages.  To keep legitimate search bots away from the trap, I set up the robots.txt rule shown above, disallowing the entire space of both legit redirector links and trap pages. Just recently, however, one of the people in the club searched Google for their own name and was quite surprised when one of the results on the first page was a link to the redirector script, with a title consisting of their e-mail address followed by my name.  Of course, they immediately e-mailed me and wanted to know how to get their address out of Google's index.  I was quite surprised too, since I had no idea that Google would index such URLs at all, seemingly in violation of my robots.txt rule. I did manage to submit a removal request to Google, and it seems to have worked, but I'd like to know why and how Google is circumventing my robots.txt like that and how to make sure that none of the disallowed pages will show up in their search results. Ps. I actually found out a possible explanation and solution, which I'll post below, while preparing this question, but  in case someone else might have the same problem.  Please do feel free to post your own answers.  I'd also be interested in knowing if other search engines do this too, and whether the same solutions work for them also. Understanding downside of triple-leveraged ETF What happens to the price of the triple-leveraged ETF if the S&amp;P 500 loses more than one-third of its value (i.e., the return is below -33 1/3%)?  Does it mean the value is equal to zero? 2 GROUP BY WITH DISTINCT , SUM , COUNT : PHP MYQL Ref :   I got answer (thanks coytech ) but i need one more column in its :  as per :  id | mid | pid | owgh | nwgh | 1    3      12    1.5    0.6 2    3      12    1.5    0.3 3    3      14    0.6    0.4 4    3      15    1.2    1.1 5    4      16    1.5    1.0 6    4      17    2.4    1.2  7    3      19    3.0    1.4   i got answer      Select mid ,          COUNT(distinct pid) as cpid  ,          SUM(nwgh) as totalnwgh  from test  GROUP BY mid   sqlfiddle :   mid  cpid       totalnwgh 3      4          3.8 4      2          2.2    But above i need one more column that's as below : totowgh  mid cpid        totalnwgh  totowgh 3      4          3.8        6.3 (DISTINCT value as per pid column) 4      2          2.2        3.9   where totowgh = 6.3 come by DISTINCT value of owgh column  as per distinct pid column  mid = 3 has count 5 but distinct pid  = 4 for mid=3 same way "distinct" owgh = 6.3 for mid=3 and distinct pid.  As pid=12 , column owgh value 1.5 is twice ...hence its is count 1 time hence,  1.5 + 0.6 + 1.2 + 3 = 6.3 ( please not this is as per DISTINCT value of pid )  Please note : i need distinct owgh value as per distinct pid or group by pid .. because if i replace value of owgh 0.6 with 1.5 then it will be 5.7 instead of 7.2 but value of owgh 0.6 belong to pid = 14 and not pid = 12 hence totalcount of owgh change ...but i need is 7.2  SEE WHAT I MEANS   : Given an alphabet of size $n$, how many strings of length $c$ contain every single letter of the alphabet at least once?  I first attempted to use a recurrence relation to work it out:  $$ T(c) = \left\{ \begin{array}{cr} 0 &amp;\mbox{ if $c&lt;n$} \\ n! &amp;\mbox{ if $c = n$} \\ T(c-1) \cdot n \cdot c &amp;\mbox{ if $c &gt; n$} \end{array} \right. $$  As there's no strings that contain every letter if c &lt; n, and if c = n then it's just all permutations. When c > n you can take any string of size (c-1) that contains all letters (of which there are $T(c-1)$ to choose from), you choose which letter to add (of which there are $n$ choices) and there are $c$ different positions to put it. However, this gives out results that are larger than $n^c$ (the total number of strings), so it can't be right, and I realised it was because you could count some strings multiple times, as you can make them taking different inserting steps.   Then I thought about being simpler: you choose n positions in the string, put each letter of the alphabet in one of those positions, then let the rest of the string be anything:  $$ {c\choose{n}} \cdot n! \cdot n^{c-n} $$  But again this counts strings multiple times.  I've also considered using multinomial coefficients, but as we don't know how many times each letter appears in the string it seems unlikely they would be much help. I've also tried several other methods, some complicated and some simple, but none of them seem to work.  How would you go about working out a formula for this? I'm sure there's something simple that I'm missing. When I import text file into R, I get a special character appended to the first value of the first column How prove $\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\ge 1$ let $a,b,c$ be postive real numbers ,and such $$ab+bc+ac=3$$   show that   $$\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge 1$$   This problem is from Secrets In Inequalities volume 1  page 30,example 1.24. the comment.the  author this case is a bit more diffcult,But this author can't post solution.Thank you I'm learning how to use the mirror modifier and I am having some trouble.  When I apply the modifier to my object I don't see a mirrored version of it.  What am I doing wrong? How to start git-gui? How to hide the [@] in a Minecraft command block Cannot see OpenLayers under Plugin in QGIS? Although I have installed the OpenLayers plugin in QGIS 2.4, I cannot see it in the list when I click on Plugins.    Does anyone else have this issue? Why didn't Palpatine publicly change his name once he became Emperor? Anakin Skywalker was known as Darth Vader (this was already his Sith name) after the creation of the Galactic Empire.  So why did Palpatine remain "Emperor Palpatine"? Why didn't he embrace his persona as Darth Sidious?  I started wondering after reading one of the answers to I have been wondering why anyone ever thought that we could find an expansion for eigenstates of some perturbed Hamiltonian in terms fo those for the basis Hamiltonian. My lecturer insisted that this was because the eigenstates of the Hamiltonian are complete', however from my (limited) understanding of the mathematical basis of QM:   When you are given a Hamiltonian, it comes with the Hilbert space of functions of which it acts (subject to some conditions we impose, such as the Lebesgue square-integrability of functions) When you are given another Hamiltonian, there is no guarantee that it will act on the same Hilbert space Therefore, that the zeroth order Hamiltonian has eigenstates forming a complete basis for the space on which ti acts does not guarantee that these same eignestates form a complete basis for the Hilbert space on which the purturbed Hamiltonian acts.   As a simple example. I can imagine adding a small perturbing magnetic field to a Hamiltonian in which there initially is none, such that we lose the degeneracy in the eigenstates accounting for different spins of a particle. One could say that we have the same Hilbert space before and after but we lose degeneracy; or I would have been tempted to say that our initial Hilbert space did not have these extra dimensions corresponding to the spin states. So rather than the eigenstates of the basis Hamiltonian also forming a complete basis for the purtubed Hamiltonian, but having eigenvalue degeneracy which is lifted on the purtubration, I would have said that our Hilbert space has changed. After all, if we take the former view (i.e. a basis accouting for spin but with degeneracy) then it seems like for any problem we would need to know and account for EVERY possible system property that can be discerned. This doesn't seem reasonable to me....  NOTE:  1) My question was asked here previously:   although there is only a discussion and no conclusive answer given. I see that the distinction is important in relativistic QM, and the case of adding a potential onto a free particle Hamiltonian seems to be a special one. Although I am still not sure what the answer is for general purturbations.  2) Similar question here:   but it seems to completely avoid answering the question. Update help center article on search to reflect the new position of the search box The page  says:      Enter search terms in the search box that appears in the upper right corner of every page    But the box is not there. Boot drops to a (initramfs) prompts/busybox caching JavaScript files Which is the best method to make the browser use cached versions of js files (from the serverside)? Let $A, B \in \mathcal M_n(\mathbb C)$, we want to know if there's any relation between their spectral radius and the spectral radius of $A + B$.  The first part of this exercise is giving examples of the three posibilities: $\rho(A) + \rho(B) &lt; \rho(A+B)$ ; $\rho(A) + \rho(B) &gt; \rho(A+B)$ and $\rho(A)+ \rho(B) =\rho(A+B)$. This part was easy because there are no conditions about A or B.  The second part is proving that if A and B are normal matrixes, (i.e. $ AA^* =A^*A$, where $A^*$ is its conjugate transpose), then $ \rho(A+B) \le \rho(A) + \rho(B) $.   So my first attempt was using the Schur's decomposition, so $A=U^{-1}D_AU$ where $D_A$ is a diagonal matrix with the eigenvalues of $A$ in its diagonal and U is an unitary matrix. Similarly, B is similar to a diagonal matrix $D_B$ with the eigenvalues of B in its diagonal. That way I have $\rho(A) = \rho(D_A)$ and $\rho(B) = \rho(D_B)$ so now since $D_A + D_B$ is a diagonal matrix with $\lambda_i + \mu_i$ for $ 1\le i \le n $, where $\lambda_i$ and $\mu_i$ are the eigenvalues of $A$ and $B$ respectively. From here it's easy to see that $ \rho(D_A+D_B) \le \rho(D_A) + \rho(D_B) $, since $ |\lambda_i+\mu_i|\le|\lambda_i|+|\mu_i| \ \forall i$. But I don't see how to include $\rho(A+B)$ in this.  And the last part of the exercise is proving that if $ \{X_i\}$ is a basis of eigenvectors, which is common to A and B, then $ \rho(A+B) \le \rho(A) + \rho(B) $.  Here's my attempt, but I would like to know if it's correct:  Let $\lambda_i$ and $\mu_i$ be eigenvalues of A and B associated to the vector $X_i$, then $AX_i=\lambda_i X_i$ and $BX_i = \mu_iX_i$, and if we sum both equalities, we have $AX_i + BX_i = \lambda_iX_i + \mu_iXi$ so $(A+B)X_i=(\lambda_i+\mu_i)X_i$ so $(\lambda_i+\mu_i)$ is an eigenvalor of $(A+B)$ associated to the eigenvector $X_i$. Then is again clear that $|\lambda_i+\mu_i|\le|\lambda_i|+|\mu_i| \le \rho(A) + \rho(B)\ \forall i$, and particularly, for $\max_i |\lambda_i+\mu_i| = \rho(A+B)$ so $ \rho(A+B)\le \rho(A) + \rho(B)$  Any kind of help is appreciated and thank you very much in advance! Evaluating $\int \frac{1}{{x^4+1}} dx$ I am trying to evaluate the integral  $$\int \frac{1}{1+x^4} \mathrm dx.$$  The integrand $\frac{1}{1+x^4}$ is a rational function (quotient of two polynomials), so I could solve the integral if I can find the partial fraction of $\frac{1}{1+x^4}$. But I failed to factorize $1+x^4$.   Any other methods are also wellcome. Before you disagree with me with a downvote, or just want to be funny, please hear me out. From the :  The upvote privilege comes first because that's what you should focus on: pushing great content to the top. Downvoting should be reserved for extreme cases. It's not meant as a substitute for communication and editing.  I have honestly never seen a downvote reserved for an extreme case. The process is usually 1) downvote post if it's bad and don't comment. (Not much of a process, is it?) As you stay on the site longer and longer, you develop a carelessness towards downvotes (as I have), but for new users, it's really discouraging.  question inspired me to [ask balpha to] make , which is a list of users with 1 rep who have the  badge. That's the closest I could get to new users who deleted their own posts after downvotes came in. Don't get me wrong, I know the argument about being nice to new users very well, but I still believe it to be an issue. And since , users are much more inclined to go the lazy route and downvote bad questions. No alternative measures (see below) are taken, for the most part. Personally, I leave a comment and don't downvote until the asker shows no interest in fixing their post. The only other measure taken to discourage downvotes (on answers) is the -1 rep punishment to the downvoter, which is absolutely meaningless and inconsequential anyways. How are we supposed to educate new users when we don't teach them what to do, don't direct them to the FAQ, and discourage them with downvotes right off the bat?  Instead of voting down: If the post is spammy or offensive — flag it. If the question is duplicate or off-topic — flag it for moderator attention. If something is wrong, please leave a comment or edit the post to correct it.  We claim to discourage downvoting, yet I see answers like  that encourage us to downvote first and ask questions later when a question is simply unclear or effortless. It's just an awful chain reaction. A new user comes to the site asking a question, but that doesn't work out too well, so they get downvotes, delete the question in shame, and don't get the help they deserve as a fellow writer of code and programmer. Strip away the quality of the site for a moment: the core functionality of this site, and the only reason I and a lot of other users are (still) on it, is to receive and give help to/from others. I enjoy making sure fellow programmers get the answer they need, because this site has helped me get mine when I needed them. Quality can still be maintained if content is fixed rather than downvoted more often. So can we stop discouraging new users from continuing to use this site and actually start to discourage downvoting? I think we need to encourage effort towards teaching and helping new users understand the system and why their post is bad rather than slapping them in the face with downvotes that they don't know how to interpret anyways. I'm leaving ways we can do this up to others. My first thought was to increase the punishment for downvoting answers and to no longer make question downvotes free, but those wouldn't be well-received, judging by the history of similar suggestions. . I would like to simulate the behaviour of a zener diode in LTspice. Consider the following schematic         &ndash; Schematic created using   I'm interested in simulating its performance when Vcc changes from the rated value (say 12V), for example when Vcc varies from 11 to 15 V.  Now, I know how to simulate this behaviour in LTspice for a generic diode or even for a particular diode which is available in the model library, but when it comes to using a custom diode, I have no idea on how to specify the diode parameters.  Looking into the documentation I was able to come up with a setup that is the correct one, that is, I can specify a custom diode model but now I'm having problems in understanding where to find all the parameters since the datasheet of my diode of choice does not provide all of them. In the LTspice model shown below I've picked a 4.7 V zener diode that was available just to have a base model to edit (and a 1k resistor just for reference).    The diode I'd like to use is a 3V . How can I simulate, as realistically as possible, the behaviour of this diode given the few parameters available in the datasheet?