Why does it take so long for the computer to respond with "invalid password", when using a correct password is near instant? When you enter a password and it's correct the response is practically instantaneous (i.e. the log on process).  When you enter an incorrect password (accidently, forgotten, etc) however, it takes a while (10-30 seconds) before it responds that the password was incorrect.  Why does it take so long (relatively) to say "incorrect password"?  This has always bugged me about entering incorrect passwords on Windows and linux (real and VM); I'm not sure about mac OSX as I can't remember if it's the same, been awhile since I last used a mac.  EDIT: for the sake of duplication, I'm asking in the context of a user logging on to the system at the physical computer rather than through ssh which could conceivably use somewhat different mechanisms to log in/validate credentials.
Why would an incorrect password attempt take a lot longer to process than a correct one? The most prominent place I've noticed this is when SSH-ing at work, but I feel like I've observed this behaviour elsewhere too.  When I try to log into Linux servers from my Windows desktop at work, I've noticed that if I mis-type my password, it takes a good 5 seconds or so before I get "Access Denied" back. Then when I type my password correctly, the login (along with welcome messages etc) is virtually instant.  Is there any logical reason for this, or would it be down to some odd configuration that's specific to the machines here at work?
Centos mysql version is 5.5 however PHPmyadmin still says version 5.1 When i run the following in my console:  [root@****~]# mysql -u root -p -e 'SELECT VERSION();' Enter password: +-----------+ | VERSION() | +-----------+ | 5.5.39    | +-----------+   Which should be the correct version.  However when i enter my PHPMYADMIN on my server it has the following specs:  Server: Localhost via UNIX socket Program: MySQL Programversion: 5.1.73 - Source distribution   So which version is the correct one and how do i make sure that the database is running on 5.5?  Note  I have already restarted sql several times without any changes Evaluation of $\displaystyle \int_{0}^{1}\frac{x\ln (x)}{\sqrt{1-x^2}}dx$   $\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{1}\frac{x\ln x}{\sqrt{1-x^2}}dx\;,$ Put $x=\cos \phi\;,$ Then $dx = -\sin \phi d\phi$  and Changing Limit, We get  $$\displaystyle I = -\int_{\frac{\pi}{2}}^{0}\cos \phi \cdot \ln(\cos \phi )d\phi = \int_{0}^{\frac{\pi}{2}} \ln(\cos \phi)\cdot \cos \phi d\phi$$  Now Using Integration by parts, We get  $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\frac{\sin^2 \phi}{\cos \phi}d\phi$$  So $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\frac{(1-\cos^2 \phi)}{\cos \phi}d\phi$$  So $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\sec \phi d\phi-\int_{0}^{\frac{\pi}{2}}\cos \phi d\phi$$  So $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\left[\ln\left|\sec \phi+\tan \phi\right|\right]_{0}^{\frac{\pi}{2}}-\left[\sin \phi\right]_{0}^{\frac{\pi}{2}}$$  So $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\left[\ln\left|\sec \phi+\tan \phi\right|\right]_{0}^{\frac{\pi}{2}}-1$$  Now How can I solve after that, Help Required, Thanks Make account merges propagate to chat I already found the radius of convergence of the power series $:$ $$\sum_{n=1}^{\infty} \frac {(-1)^n} {n} z^{n(n+1)}.$$ What happens at the boundary of disk of convergence?  I found that the radius of convergence to be $1$ and hence the disk of convergence is $|z|&lt;1$. Now I found at $z=1$ the series converges. Now how can I proceed? Please help me. Thank you in advance. Auto.arima with daily data: how to capture seasonality/periodicity? I am fitting an ARIMA model on a daily time series. Data are collected daily from 02-01-2010 to 30-07-2011 and are about newspaper sales. Since a weekly pattern in sales can be found (the daily average amount of copies sold is usually the same from Monday to Friday, then increases on Saturday and Sunday), I am trying to capture this "seasonality". Given the sales data "data", I create the time series as follows:  salests&lt;-ts(data,start=c(2010,1),frequency=365)   and then I use the auto.arima(.) function to select the best ARIMA model via AIC criterion. The result is always a non-seasonal ARIMA model, but if I try some SARIMAs model with the following syntax as example:  sarima1&lt;-arima(salests, order = c(2,1,2), seasonal = list(order = c(1, 0, 1), period = 7))   I can obtain better results. Is there anything wrongs in the ts command / arima specification? The weekly pattern is very strong so I would not expect so many difficulties in capturing it.  Any help would be very useful. Thank you, Giulia Deppieri  Update:  I have already changed some arguments. More precisely, the procedure selects ARIMA(4,1,3) as the best model when I set D=7, but AIC and the others good of fit indexes and forecasts as well) do not improve at all. I guess there's some mistakes due to confusion between seasonality and periodicity..?!   Auto.arima call used and output obtained:  modArima&lt;-auto.arima(salests,D=7,max.P = 5, max.Q = 5)     ARIMA(2,1,2) with drift         : 1e+20  ARIMA(0,1,0) with drift         : 5265.543  ARIMA(1,1,0) with drift         : 5182.772  ARIMA(0,1,1) with drift         : 1e+20  ARIMA(2,1,0) with drift         : 5137.279  ARIMA(2,1,1) with drift         : 1e+20  ARIMA(3,1,1) with drift         : 1e+20  ARIMA(2,1,0)                    : 5135.382  ARIMA(1,1,0)                    : 5180.817  ARIMA(3,1,0)                    : 5117.714  ARIMA(3,1,1)                    : 1e+20  ARIMA(4,1,1)                    : 5045.236  ARIMA(4,1,1) with drift         : 5040.53  ARIMA(5,1,1) with drift         : 1e+20  ARIMA(4,1,0) with drift         : 5112.614  ARIMA(4,1,2) with drift         : 4953.417  ARIMA(5,1,3) with drift         : 1e+20  ARIMA(4,1,2)                    : 4960.516  ARIMA(3,1,2) with drift         : 1e+20  ARIMA(5,1,2) with drift         : 1e+20  ARIMA(4,1,3) with drift         : 4868.669  ARIMA(5,1,4) with drift         : 1e+20  ARIMA(4,1,3)                    : 4870.92  ARIMA(3,1,3) with drift         : 1e+20  ARIMA(4,1,4) with drift         : 4874.095   Best model: ARIMA(4,1,3) with drift           So I assume the arima function should be used as:  bestOrder &lt;- cbind(modArima$arma[1],modArima$arma[5],modArima$arma[2]) sarima1&lt;-arima(salests, order = c(4,1,3))   with no seasonal component parameters and period specifications. Data and exploratory analysis show that the same weekly pattern can be approximatively considered for each week, with the only exception of August 2010 (when a consistent increase in sales is registered). Unfortunately I have no expertise in timeseries modeling at all, in fact I am trying this approach in order to find an alternative solution to other parametric e non-parametric models I have tried to fit for these problematic data. I have also many dependent numeric variables but they have shown low power in explaining the response variable: undoubtedly, the most difficult part to model is the time component. Moreover, the construction of dummy variables to represent months and weekdays turned out not to be a robust solution. How to create a good glass reflection with Principled BSDF Show how if Alice uses the same value of $k$ to sign two different messages $m_1$ and $m_2$, using the ElGamal signature scheme, Eve can recover the value of $a$ from the corresponding signatures $(m_1, r_1, s_1)$ and $(m_2, r_2, s_2)$. (Note: you are allowed to assume that if $\gcd(a, n) = d$ then there are $d$ solutions to the congruence $ax \equiv b \pmod n$.)     ElGamal Signature Scheme:      Key Gen:      Compute $y = g^x \;\bmod p$.      The public key is $(p, g, y)$. The secret key is $(p, g, x)$      Signature Gen:      Choose a random $k$ such that $0 &lt; k &lt; p − 1$ and $\gcd(k, p − 1) = 1$      Compute $r = g^k \;\bmod p$ and $s = k^{-1}(m – xr) \;\bmod{p-1}$   Thoughts: So far, I can tell that $r_1$ and $r_2$ are equal and $s_1$ and $s_2$ are closely related since the only variation is $m$.  We can relate the two equations for $s$, by solving them for $-xr$:  $$s_1k - m_1 \equiv s_2k - m_2 \pmod p$$  $$(s_1 - s_2)k \equiv m_1 - m_2 \pmod p$$  Let $a = s_1 - s_2$, then we know from the question that for $\gcd(a, n) = d$ there are $d$ solutions for $k$.  Now the forger can compute $g^k$ for each solution to k, until $r$ is found.  Then compute $xr \equiv m_1 - ks_1 \pmod{p-1}$  There are $d&#39; = \gcd(r, p-1)$ solutions for $x$.  The forger can compute $g^x$ for all the $x$s found until she finds $y$.  Once she has $y$ she knows the proper $x$ which can be used to find $m_1$ and $m_2$.  How does this look? Clock shows wrong time after switching from Ubuntu to Windows 10 I have dual booted Windows 10 and Ubuntu 16.04 in UEFI mode (secure boot off). My time in BIOS is correct.  When I boot into Ubuntu and then boot into Windows 10, it shows the wrong time. I have to correct it manually or by the internet in Windows 10, and when I reboot into Windows 10 then it is not a problem. The clock will show the correct time. And then when I reboot to Linux the clock is correct, but switching back to Windows 10 the clock messes up again. I want to display the pages 'famous' and 'mschief'.  Using this code, WP_query() doesn't return any result:  $args = array( 'post_type' =&gt; 'page', 'pagename' =&gt; 'famous,mschief', 'posts_per_page' =&gt; 3 );   This code returns one of the 2 pages:  $args = array( 'post_type' =&gt; 'page', 'pagename=famous,mschief', );   And this code returns the 2 pages correctly:  $args = array( 'post_type' =&gt; 'page', 'pagename=famous,mschief', 'posts_per_page' =&gt; 3 );   Why does the first piece of code (the only correct one to me) doesn't work? And why the two others produce these results? Integral ${\frac{1}{\pi^2}} \int_{0}^{\infty}\frac{{(\ln{x}})^2}{\sqrt{x}{(1-x)^2}} \mathrm d x$ Number of vertices present in all maximum matchings Given a graph $G$, we need to find the cardinality of the largest set of vertices so that each of them are present in every maximum matching possible.   Is there a solution beside the obvious remove each vertex and find the maximum matching to see it reduces? A function $f: \mathbb R \to \mathbb R$ satisfies equation $f(x+y)=f(x)f(y)$ for all $x,y\in \mathbb R $ Accoring to Joseph Wright's answer at   The k-type argument has recently been adjusted: you now want e-type  Questions:  What is the functionality of the new e-type argument and how does it differ from the previous k-type argument?  The 2016-11-21 version of the xparse documentation does not mention the k-type argument at all. Is the plan to leave the earlier k functionality intact (but deprecated so no documentation) or will it eventually be removed and any usage of that should eliminated? If the plan is for it to be removed, how long will the older k-type argument functionality be kept in xparse.   Code: Example of the k-type argument from . Would be useful to see how this would be modified to use the new e-type argument. \documentclass{article}  \usepackage{amsmath} \usepackage{xparse} \usepackage{xcolor}  \NewDocumentCommand\MyMacro{k_}{%     \IfNoValueTF{#1}{%         \mathbf{A}     }{%         A_{\textcolor{red}{#1}}     }% }  \begin{document}  With a subscript: $\MyMacro_{\pi}$   Without any subscript: $\MyMacro$   \end{document} I  just got these 10 e-mails 5 hours ago from do-not-reply@stackexchange.com:      Recently, you asked a question on [X] Stack Exchange, but you haven't verified your email address yet.      Click on the following link to verify your email and set a password so that you can log in.      Here, [X] refers to all of the following:    GameDevelopment.SE SharePoint.SE Unix.SE  Skeptics.SE Japaneese.SE HomeImprovement.SE SciFi.SE PersonalFinanceMoney.SE  DatabaseAdmins.SE  TeX.SE    I haven't posted on any of these except for TeX.SE, but that was not recently, and through my real account.   =   I asked about this , and was told that someone seems to be impersonating me by using my email. Is it possible for staff to stop this?   Also, I'm pretty surprised there is no measure to stop people from using emails to existing SE accounts (see for example, on wordpress.com, if you use an e-mail which has an existing account on wp.com, you are asked to log in.).    EDIT:  I just got 2 more emails an hour ago for SharePoint.SE, and for AmateurRadio.SE. Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$. Is the following true? Let $x_n$ be a sequence with the following property: Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.  I guess that it is true but I am not sure how to prove this. Schengen area 90 days limit: Can I leave on the 91st day? Can anyone tell me if the 90 days is in fact 90 nights? Can you leave the Schengen area on the 91st day? Or are you only allowed 89 nights and you are required to leave on the 90th day? When I look at the definition given from the Mac OS X Dictionary (I have set American English as interface language, and the dictionary used is then the New Oxford American Dictionary), I read:     the /ði/ /ð(ə)/ /ði/ [called the definite article]   adjective   1. denoting one or more people or things already mentioned or assumed to be common knowledge: what's the matter? | call the doctor | the phone rang. Compare with a.   • used to refer to a person, place, or thing that is unique: the Queen | the Mona Lisa | the Nile.   • informal denoting a disease or affliction: I've got the flu.   • (with a unit of time) the present; the current: dish of the day | man of the moment.   • informal used instead of a possessive to refer to someone with whom the speaker or person addressed is associated: I'm meeting the boss | how's the family?   • used with a surname to refer to a family or married couple: the Johnsons were not wealthy.   • used before the surname of the chief of a Scottish or Irish clan: the O'Donoghue.   I have never thought of an article as adjective. Is normally an article defined as adjective? I need to amend my return but I already received check from IRS with my refund. Should I cash it out now or wait until I amend my return? I need to amend my return, but I have already received a check from IRS (they approved my return that I need to amend). Should I cash out this check and then amend my return, or should I dispose this check and wait for a check from an amended return? Any good tutorials for learning how to post-process images? I don't really know where to start with post-processing my images. I normally just straighten, crop and sharpen. Do people know of any good resources on the internet for getting a grip with Photoshop and similar tools? Is verbatim copying several paragraphs of text with citation considered plagiarism? Is it plagiarism if I copy several paragraphs from another's source (let's say a CS paper) into my work and then footnote Source: ..., for example, as background information?