My /User directory permissions always changes… should I be worried?
Permissions of /Users keep changing back to drwxrwxrwx
Why does space expansion not expand matter? I have looked at other questions on this site (e.g. "why does space expansion affect matter") but can't find the answer I am looking for.  So here is my question: One often hears talk of space expanding when we talk about the speed of galaxies relative to ours. Why, if space is expanding, does matter not also expand? If a circle is drawn on balloon (2d plane), and the balloon expands, then the circle also expands. If matter is an object with 3 spatial dimensions, then when those 3 dimensions expand, so should the object.  If that was the case, we wouldn't see the universe as expanding at all, because we would be expanding (spatially) with it.  I have a few potential answers for this, which raise their own problems:   Fundamental particles are 'point sized' objects. They cannot expand because they do not have spatial dimension to begin with. The problem with this is that while the particles would not expand, the space between them would, leading to a point where the 3 non-gravity forces would no longer hold matter together due to distance Fundamental particles are curled up in additional dimensions a la string theory. These dimensions are not expanding. Same problems as 1, with the added problem of being a bit unsatisfying. The answer seems to be (from Marek in the previous question) that the gravitational force is so much weaker than the other forces that large (macro) objects move apart, but small (micro) objects stay together. However, this simple explanation seems to imply that expansion of space is a 'force' that can be overcome by a greater one. That doesn't sound right to me. I have a model similar to below with a continuous outcome, continuous predictor and a categorical variable. the goal of this analysis is to make inferences on the group variable. The issue is that y is bounded by x in the form of y >= -x. When the model makes predictions you can see in the plot below that the predicted value is outside the possible range. See the last plot.   How can this be remedied? A custom loss function to penalize going beyond the bound and use gradient descent regression? Or is there some other way? Thank you.   x =   rep(seq(-10,10,1),each=5)  y =  rep(0,length(x) ) weights = sample( seq(1,1000,1) ,length(x)) groups = rep( letters[1:5], times =length(x)/5  )   ## the outcome y has the relationship y &gt;= -x for(i in 1:length(x)){   y[i] = sample(  seq(-x[i],-x[i]+ ifelse(x[i]&lt;0 ,                                            sample(seq(1,10,.1),1),                                           sample(seq(5,40,.1),1) ),1)                        ,1) }   dat =data.frame(x = x, y = y, weights = weights, group = groups) model = lm(dat = dat, y ~ x + group + group * x, weights = weights) dat$FIT= predict(model, dat, allow.new.levels = TRUE)  library(ggplot2) ggplot(data = dat, aes(x = x, y = y, color = group))+geom_point( aes(size = weights))+ geom_line(data = dat, aes(x = x, y = FIT, color = group))    #data for new predictions newdata = data.frame( x = seq(-100,100,1) ,                        group = rep("c", length( seq(-100,100,1) )),                       bound = -seq(-100,100,1)                       )  ## when pridicting new data the outcome y has the relationship y &gt;= -x ## but the predicted vales are below the bound. See how red line is below blue to the left of zero new_preds  = predict(model, newdata, allow.new.levels = TRUE) ggplot(data = newdata, aes(x = x, y = new_preds, color = group))+geom_line()  +   geom_line(data = newdata, aes(x = x, y = bound, color = "green")) Change System Language Solaris 8 I have Solaris 8 OS installed in French.  Is it possible to switch from French to English without reinstalling the OS ? If yes how ?  When locale -a i have this :  POSIX common en_US.UTF-8 C iso_8859_1 en_CA en_CA.ISO8859-1 en_US en_US.ISO8859-1 en_US.ISO8859-15 en_US.ISO8859-15@euro fr_CA fr_CA.ISO8859-1 iso_8859_15 th th_TH th_TH.TIS620   Even when I try to edit /etc/default/init with :  LANG=en_US.ISO8859-15 LC_COLLATE=en_US.ISO8859-15 LC_CTYPE=en_US.ISO8859-15 LC_MESSAGES=en_US.ISO8859-15 LC_MONETARY=en_US.ISO8859-15 LC_NUMERIC=en_US.ISO8859-15 LC_TIME=en_US.ISO8859-15 LC_ALL=en_US.ISO8859-15   and reboot : nothing happens.  To verify the result i always use date and when i reboot : i also have the date in french.  P.S : I have to do this configuration in a computer which isn't here in my office, so i have to this manipulation in a an another computer here in my office and if it works, i'll do it to the other computer.  So the computer here is configured in English and i have to switch it to French. Here is the content of my ~./dt :  C                     errorlog              sessionlogs Desktop               errorlog.old          sessions Trash                 errorlog.older        simulateur.dtwmrc appconfig             help                  startlog appmanager            icons                 startlog.old asr.dtwmrc            outils.dtwmrc         startlog.older bin                   palettes              tmp bookmarks             sdtfonts              types dtwmrc                sdtpcaction.conf      wsmenu en_US.ISO8859-15      sdtpcaction.conf.bak   Thanks How can I switch between starting a coroutine vs running the routine all at once? I'm generating a level, and I have an optional delay to show the steps level generation one by one.  If my generationDelay boolean is true, I want to use StartCoroutine to spread the generation out over time. If it's false, I want to call the same method but without StartCoroutine.  The problem is that the method is type IEnumerator.  I can make two methods one type of IEnumerator and one not. But I wonder if there is a way to use one method.  In the first script top i did:  public bool generationDelay = false;   Then in a method:  private void BeginGame()     {         mazeInstance = Instantiate(mazePrefab) as Maze;          if (generationDelay == true)         {             StartCoroutine(mazeInstance.Generate(generationDelay));         }         else         {             mazeInstance.Generate(generationDelay);         }     }   Then the other script where the method Generate is: In the top:  public float generationStepDelay = 0.01f;   Then the method Generate:  public IEnumerator Generate (bool generationDelay) {          if (generationDelay == true)             WaitForSeconds delay = new WaitForSeconds(generationStepDelay);         cells = new MazeCell[size.x, size.z];         List&lt;MazeCell&gt; activeCells = new List&lt;MazeCell&gt;();         DoFirstGenerationStep(activeCells);         while (activeCells.Count &gt; 0) {             yield return delay;             DoNextGenerationStep(activeCells);         }     }   I'm getting an error on the line:   WaitForSeconds delay = new WaitForSeconds(generationStepDelay);      "Embedded statement cannot be a declaration or labeled statement"   And an error on the line:  yield return delay;      "The name 'delay' does not exist in the current context"   The reason I want to use a bool variable is that when i'm using StartCoroutine it's kind of slow, even if I change the value of generationStepDelay to 0 it's kind of slow. And if I'm not using StartCoroutine and not using the Generate method as IEnumerator it will work fast.  So i wonder what should I do and how? Live notification shown for own edits When editing an answer, the &quot;new revision&quot; message is shown for a short moment. This makes no sense because the revision you just created is automatically displayed anyway. Using the properties of real numbers, verify commutativity on complex numbers. I have just started self-studying Linear Algebra after a 3 year hiatus from mathematics (in an extended neuroscience foray), and I want to ensure that I am sufficiently verifying commutativity of complex numbers by using the properties of real numbers.  My attempts...  (1) Commutativity of Addition:   Verify that $w + z = z + w$ for all $w, z \in \mathbb{C}$.  $$w + z = (a + bi) + (c + di) \hspace{10mm}\text{by definition of }\mathbb{C}$$ $$\hspace{13mm} = a + (bi + c) + di \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{17mm} = a + (c + bi) + di \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{16mm} = (a + c) + (bi + di) \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{19mm} = (c + a) + (di + bi) \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{13mm} = c + (a + di) + bi \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{17mm} = c + (di + a) + bi \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{17mm} = (c + di) + (a + bi) \hspace{11mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{13mm} = z + w \hspace{34mm}\text{by definition of }\mathbb{C}$$  $$\blacksquare$$  This seems a bit verbose to me - is it acceptable to combine steps? For example, rather than:  $$w + z = (a + bi) + (c + di) \hspace{10mm}\text{by definition of }\mathbb{C}$$ $$\hspace{13mm} = a + (bi + c) + di \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{17mm} = a + (c + bi) + di \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ can I instead write: $$w + z = (a + bi) + (c + di) \hspace{10mm}\text{by definition of }\mathbb{C}$$ $$\hspace{48mm} = a + (c + bi) + di \hspace{10mm}\text{by associativity and commutativity in }\mathbb{R}$$?  (2) Commutativity of Multiplication:  Verify that $wz = zw$ for all $w, z \in \mathbb{C}$.  $$\hspace{-14mm}wz = (a + bi)(c + di) \hspace{24mm}\text{by definition of }\mathbb{C}$$ $$\hspace{4mm} = a \cdot c + a \cdot di + bi \cdot c + bi \cdot di \hspace{10mm}\text{by distribution in }\mathbb{R}$$ $$\hspace{9mm} = a \cdot c + bi \cdot c + a \cdot di + bi \cdot di \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{35mm} = c \cdot a + c \cdot bi + di \cdot a + di \cdot bi \hspace{10mm}\text{by multiplicative commutativity in }\mathbb{R}$$ $$\hspace{1mm} = c(a + bi) + di(a + bi) \hspace{20mm}\text{by distribution in }\mathbb{R}$$ $$\hspace{5mm} = (c + di)(a + bi)\hspace{34mm}\text{by distribution in }\mathbb{R}$$ $$\hspace{4mm} = zw \hspace{55mm}\text{by distribution in }\mathbb{R}$$ $$\blacksquare$$  This also seems a bit wonky to me. Any and all suggestions on clearing up my verification, reasoning, etc. would be greatly appreciated! Prove that $A+I$ is invertible if $A$ is nilpotent Possible Duplicate:          Given $A^{2012}=0$ prove that $A+I$ is invertible and find an expression for $(A+I)^{-1}$ in terms of $A$. ($I$ is the identity matrix). In the sentence: "Drinking and driving IS dangerous", why does the verb "to be" need to be singular? In the sentence: "Drinking and driving is dangerous". Drinking and driving are both subjects and the word dangerous is the object (I think) but why do I have to use the verb to be in singular? The number of divisors I am reading a math book. It states the following rule:      For an integer $n$ greater than 1, let the prime factorization of $n$ be $$n=p^a_1p^b_2p^c_3...p^m_k$$    Here $a, b, c,..., m$ are nonnegative integers, and $p_1,p_1,...,p_k$ are prime numbers. Then the number of divisors of $n$ is: $$d(n)=(a+1)(b+1)(c+1)...(m+1)$$   I can use the rule to solve problems correctly, but I don't understand the rule. How to prove it? Why didn't Harry Potter tell Aunt Petunia he was rich to improve how he was treated by Dursleys? This is a follow-up question to   One reason Aunt Petunia mistreated Harry was that she did not know he was rich. If she knew, she would naturally have treated this golden goose nicely. Her retirement would be secured if Harry financially supported her in old age.  Why didn't Harry tell Aunt Petunia he was rich? Surely every boy wants to be treated nicely by his family. I've downloaded the new iOS 8 update on my iPhone 5 but am getting second thoughts about actually installing it.  Is there a way for me to delete that download or am I at the point of no return? Accidental chown under / as root I accidentally run the following command as root chown mike -Rf / backups the point is that I added a space between / and backups my mistake, I stopped the command after 1-2 seconds, how can I recover from this? I have problems with DirectAdmin right now for example when accessing it I recive the error:  Unable to determine Usertype user.conf needs to be repaired http://help.directadmin.com/item.php?id=456   Or I stopped recieving emails, I dont even know what else is messed up right now... A question about the arctangent addition formula. In the arctangent formula, we have that:  $$\arctan{u}+\arctan{v}=\arctan\left(\frac{u+v}{1-uv}\right)$$  however, only for $uv&lt;1$. My question is: where does this condition come from? The situation is obvious for $uv=1$, but why the inequality?  One of the possibilities I considered was as following: the arctangent addition formula is derived from the formula:  $$\tan\left(\alpha+\beta\right)=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}.$$  Hence, if we put $u=\tan{\alpha}$ and $v=\tan{\beta}$ (which we do in order to obtain the arctangent addition formula from the one above), the condition that $uv&lt;1$ would mean $\tan\alpha\tan\beta&lt;1$, which, in turn, would imply (thought I am NOT sure about this), that $-\pi/2&lt;\alpha+\beta&lt;\pi/2$, i.e. that we have to stay in the same period of tangent.  However, even if the above were true, I still do not see why we have to stay in the same period of tangent for the formula for $\tan(\alpha+\beta)$ to hold. I would be thankful for a thorough explanation. How can I know when my screen was locked last time? Exist a log where I can look for that information? I want to know last days when my PC automatically locked the screen because was idle. Prevent Mac Terminal from reverting to default theme I can't find any way to change the Terminal theme and have that theme applied permanently.  I want to change to another darker theme other than the default 'basic' theme, and (most importantly) have it remain as the theme every time I open Terminal. Show $f$ can't be irreducible over a finite field if $f^\prime$ is the zero polynomial. Shoot, Shoot At I have a question about the difference between the verb "shoot" and "shoot at":          Police shot the suspect.     Police shot at the suspect.        Are they just the same? windows 7 desktop icons: new icon placement When moving/saving/etc a new file on the desktop, the icon appears second (just after "Computer"). I find it annoying, since I want my "special" icons to be first (such as the recycle bin, "Network", etc.).  Can this behavior be changed? Quick question about curly braces not showing up Ii am writing a short code, so I doubt anything else in the code is causing the problem. This is the problematic line of code:  \[ M_R := \max {|f(z)|: |z - z_0| = R }\le M\]   It doesn't show the curly braces. I've tried using \left{ and \right} but to no avail.  What should I do?  Thank you The vast majority of intelligent species in the Star Trek universe are bipedal, humanoid creatures.  From the lot I've seen of TOS, TNG, DS9, Voyager, Enterprise, and the movies, it seems that this coincidence goes largely unquestioned.  Metaknowledge grants that it is much easier to have human actors portray human-like beings, and so this was probably done due to budgetary/technological limitations.  However, I'm wondering if an in-universe reason for this was ever given or explored?