I was just thinking that maybe if we have more Stack&nbsp;Exchange websites then each person would be a member of more different sites on average. Since the time a person can spend online on average is limited it means he will be spending less time on each website. But would this affect the quality of these websites? I'm not talking only about the quality answers given but also the number of answers given, the promptness. Won't these things go down?  I think that the rate of questions won't go down much because you need to know what you need to know. However, the average time a person spends browsing through these websites (and chances upon a question he can answer) would decrease. So are we creating too many different Stack&nbsp;Exchange sites?
have already been created through .  have already completed their beta stage and are waiting for launch.  are in public beta stage.  are in commitment phase in   proposals are listed in   Now it is really confusing to find a suitable site to ask a question. As an example - I host a blog using . Now I have a question about this hosting. I want to post it in a . Where will I post? I have 2 options available now.     Are developers trying anything to solve this confusing matter?
Simpler proof - Non atomic measures Suppose that $(X,\mathcal{E},\mu)$ is a non-atomic finite measure space (i.e. for every $E \in \mathcal{E}$ with $\mu(E)&gt;0$ there exists $F \subset E$ measurable such that $0&lt;\mu(F) &lt;\mu(E)$.)  a) Prove that for every $ \varepsilon &gt;0$ there is a finite partition of $X$ in measurable subsets $X_1,..,X_n$ such that $\mu(X_i)\leq \varepsilon$.  b) Prove that for every $\alpha \in [0,\mu(X)]$ there exists $E \in \mathcal{E}$ with $\mu(E)=\alpha$.    I guess that a) is given to prove b) more easily. I have the following idea of solution for a) (inspired from , which proves b)  )  Denote  $$ \Gamma = \{ (X_1,..,X_n) : n &gt;&gt; \mu(X)/\varepsilon, X_i \text{ are disjoint }, \mu(X_i)\leq \varepsilon \}$$ ordered by componentwise inclusion. Totally ordered parts $(Y_\alpha)$ of $\Gamma$ have an upper bound element $(\bigcup Y_\alpha^i)_{i=1}^n$ which is still in $\Gamma$.   By Zorn's lemma $\Gamma$ has maximal elements. If a maximal $(X_1..X_n)$ element is not a partition, then we can replace it with something better, of the form $(X_1,..,X_n\cup A)$ where $A \subset X \setminus (X_1 \cup..\cup X_n)$ and $\mu(A)&gt;0$.     My questions are:  1) Is my solution of a) correct? I feel that the part with the upper bound element may not work, since there might be noncountable unions.  2) What is a simpler solution of a)? Let $R$ be a ring, where $a^{3} = a$ for all $a\in R$. Prove that $R$ must be a commutative ring. In C, how do I restrict the scope of a global variable to the file in which it's declared? I have a directory that contains the following:  x.pdf y.zip z.mp3 a.pdf   I want to delete all files apart from x.pdf and a.pdf. How do I do this from the terminal? There are no subdirectories so no need for any recursion. What is the limit of this sequence? How does it relate to the exponential function? Find $$\lim_{n\to\infty} \left(1+\frac{x^2}{n^2}\right)^n$$   I particular, I am hoping to find the above to be $1+x^kg(x/n)$ where $g(\cdot)$ is a function with uniformly bounded derivatives.  Edit: I'm trying to prove a result that would be true if  $(1+x^2/n^2)^n -1 = x^kO(x/n).$ Print box drawing characters with pdfLaTeX I need to print some unicode characters (of box drawing family) in my document. I found this  that is very similar to mine but is valid only for XeLaTeX. There is a valid way to do this with pdfLaTeX or I have to use XeLaTeX instead? I am trying to create Thiessen (Voronoi) polygons, based on points within a defined shapefile extent. The output should be inclusive to the feature.  The output I receive now is the rectangular area around the selected points.    Steps I am using: 1. select points of interest (inside and outside spatial extent) 2. set primary display field to FID (aka:ObjectID) 3. Environment… > Set 'Analysis Mask' to shapefile with feature (in this case, a watershed)  Now here is where the problem starts. I have a set of instructions for arc 9.3, but I am now operating on ArcGIS 10 (Spatial Analyst installed &amp; activated). Also, just to clarify, I did scour the inter-webs for some number of hours- Esri support, gis.SE, Professor Google...  ########  9.3.1 version: Spatial Analyst >> Distance >> Allocation - set 'Assign To' as point shapefile of selected data - set cell size  10.0 methods tried: Analysis >> Create Thiessen Polygons -- Many variations of selected features -- File locations, both inside and out of file geodatabase  I am open to any solution, using Arc 10.0 and/or python would be ideal.  The first image is an example.  The output does not need to be exact.  wiki: In (non-relativistic) classical physics, if the temperature of an object is proportional to the average kinetic energy ${1 \over 2} m\overline {v^{2}}$of its particles (or molecules), then shouldn't that temperature depend on the frame of reference - since $\overline {v^{2}}$ will be different in different frames?  (I.e. In the lab frame  $K_l = {1 \over 2} m\overline {v^{2}} $, but in a frame moving with velocity $u$ relative to the lab frame, $K_u =  {1 \over 2} m \overline {(v+u)^{2}}$). How should I parse the sentence "Buffalo buffalo Buffalo buffalo buffalo buffalo Buffalo buffalo." How to disable sticky corners in Windows 10 It seems Microsoft has tried to solve this problem .  In Windows 7, there is no boundary between monitors and the mouse can move freely across the top of the screen between my 3 monitors, provided I had set their heights equal in display settings. Windows 8 introduced sticky corners. See .  Windows 10 got sticky corners as well. There are a few pixels at the top corners of each display where the mouse cannot cross over onto the other display. One must move the cursor down to avoid this region in order to get to the next display.  The image shows roughly the region where mouse movement is not permitted in Windows 10, but is allowed in Windows 7.    Personally, I had no problem with unrestricted mouse movement across the top of my screens - I got used to "aiming" for the x, and the convenience of unrestricted cursor movement. Like all the people who wanted to disable it in W8, I'm wondering if there is a way to disable it in W10.  Edit to address possible duplicate:  Although the problem is identical to the one in , solutions to solve the problem in W8 involving editing the registry key MouseCornerClipLength do not work in W10, since that registry key is not present in W10. Also adding that key and setting the value doesn't work. I searched the entire registry and couldn't find it in another location. No other W10 keys in the node referenced in the W8 solution are obvious replacements.  Edit to address possible solutions in comments  harrymc's  and  that worked for for Windows 8.1 do not work in Windows 10. How do you extract IP addresses from files using a regex in a linux shell? Goofy looking alien robots lose a "Kleptonite Ball" towed behind their spacecraft Let $r$ be the smallest positive integer such that $a^r \equiv 1 \bmod{n}$. Prove $r \mid \phi(n)$. Let $a \in \mathbb {Z}$, $n &gt; 1$ a natural number with $\gcd(a, n) = 1$, and let $r$ be the smallest positive integer such that $a^r \equiv 1 \bmod{n}$. Prove that $r \mid \phi(n)$.  Euler's generalization of Fermat's Little Theorem: If $n \in \mathbb {N}$ and $m \in \mathbb{Z}$ such that $\gcd(m, n) = 1$, then $m^{\phi(n)} \equiv 1\bmod(n)$.  By Euler's generalization I would have $a^{\phi(n)} \equiv 1\bmod{n}$.  Then I could say $a^r \equiv a^{\phi(n)} \equiv 1 \bmod{n}$. At this point I can see that $r$ clearly must be some multiple of $\phi(n)$.  If I want to show that $r \mid \phi(n)$, then I can show that $\phi(n) = rk$, for some $k \in \mathbb {Z}$.  However, I am a bit stuck about how to make this last step showing that  $\phi(n) = rk$, where $k \in \mathbb {Z}$. How can you dynamically create variables via a while loop? Estimating partial sums $\sum_{n = 1}^m \frac{1}{\sqrt{n}}$ Apostol's Calculus, exercise number I 4.7 13.     Prove that if $n \geq 1$, then   $$ 2(\sqrt{n+1} - \sqrt{n}) &lt; \frac{1}{\sqrt{n}} &lt; 2(\sqrt{n} - \sqrt{n-1}) $$   and use this to prove that if $m \geq 2$, then   $$ 2 \sqrt{m} - 2 &lt; \sum_{n = 1}^m \frac{1}{\sqrt{n}} &lt; 2\sqrt{m} - 1 $$   In particular, when $m = 10^6$ the sum lies between $1998$ and $1999$.   After establishing the trivial case, for $n=1$, I can't think of a way to perform the inductive step:  assuming $2(\sqrt{n+1} - \sqrt{n}) &lt; \frac{1}{\sqrt{n}} &lt; 2(\sqrt{n} - \sqrt{n-1})$ and using this, deducing the same for $n+1$ is true as well:  $2(\sqrt{n+2} - \sqrt{n+1}) &lt; \frac{1}{\sqrt{n+1}} &lt; 2(\sqrt{n+1} - \sqrt{n})$  Any help is very appreciated, thanks Compare the statistical significance of the difference between two polynomial regressions in R So first of all I did some research on this forum, and I know   questions have been asked but they usually haven't been answered properly or sometimes the answer are simply not detailed enough for me to understand. So this time my question is : I have two sets of data, on each, I do a polynomial regression like so :  Ratio&lt;-(mydata2[,c(2)]) Time_in_days&lt;-(mydata2[,c(1)]) fit3IRC &lt;- lm( Ratio~(poly(Time_in_days,2)) )   The polynomial regressions plots are:    The coefficients are :  &gt; as.vector(coef(fit3CN)) [1] -0.9751726 -4.0876782  0.6860041 &gt; as.vector(coef(fit3IRC)) [1] -1.1446297 -5.4449486  0.5883757    And now I want to know, if there is a way to use an R function to do a test that would tell me whether or not there is a statistical significance in the difference between the two polynomials regression knowing that the relevant interval of days is [1,100].  From what I understood I can not apply directly the anova test because the values come from two different sets of data nor the AIC, which is used to compare model/true data.  I tried to follow the instructions given by @Roland in the related question but I probably misunderstood something when looking at my results :  Here is what I did :   I combined both my datasets into one.  f is the variable factor that @Roland talked about. I put 1s for the first set and 0s for the other one.  y&lt;-(mydata2[,c(2)]) x&lt;-(mydata2[,c(1)]) f&lt;-(mydata2[,c(3)])  plot(x,y, xlim=c(1,nrow(mydata2)),type='p')  fit3ANOVA &lt;- lm( y~(poly(x,2)) )  fit3ANOVACN &lt;- lm( y~f*(poly(x,2)) )   My data looks like this now :    The red one is fit3ANOVA which is still working but I have a problem with the blue one fit3ANOVACN the model has weird results. I don't know if the fit model is correct, I do not understand what @Roland meant exactly.  Considering @DeltaIV solution I suppose that in that case :  The models are significantly different even though they overlap. Am I right to assume so ? What's the difference between mobile and desktop processors? Proof of upper-tail inequality for standard normal distribution $X \sim \mathcal{N}(0,1)$, then to show that for $x &gt; 0$, $$ \mathbb{P}(X&gt;x) \leq \frac{\exp(-x^2/2)}{x \sqrt{2 \pi}} \&gt;. $$ I uploaded a pre-print of an article on arXiv 1 year ago. The article was accepted in SCI journal and currently available online. However, I found a typo mistake in one equation and want to add more discussion related one figure (On researchgate I am receiving a lot of queries regarding that figure). Should i upload a new version of pre-print on arXiv? (How citations will work in this case? as arxiv version is more advanced compare to Journal version) OrI may upload a short document on arXiv as an errata/correction with extra discussion?Or I may address errata and discussion in online page of journal? Actually this journal allows the registered users to comment and start discussion related to published article. Get network utilization from command line In the Windows task manager there is a handy network tab which shows the link speed and network utilization (in percentage of available bandwidth) for connected network adapters:  Is there a way to get this information through the command line?