What is the probability that a random $\{0,1\}$, $n \times n$ matrix is invertible? Assume the 0 and 1 are each present in an entry with probability $\frac{1}{2}$. Is there an explicit formula as a function of $n$?  Does it tend to 1 as $n$ grows large? I'm sure this is all known...  Thanks!
Given a random square matrix of size $n\times n$ in the field $\mathbb F_2$, what is the probability that its determinant is $1$? (This is also the probability that the matrix is non-singular, since $\mathbb F_2$ only has the elements $0$ and $1$.)
How do you do load testing and capacity planning for web sites? This is a  about capacity planning for web sites. Related:      What are some recommended tools and methods of capacity planning for web sites and web-applications? Please feel free to describe different tools and techniques for different web-servers, frameworks, etc., as well as best-practices that apply to web servers in general. On my mac I have 2 accounts and the both are normal account. So in this moment I don't have a Administrator account, only one that was in the past a administrator, but now is not.  How I can have again a Administrator account? If $u$ is harmonic and bounded in $0 < |z| < \rho$, show that the origin is a removable singularity This is a reworking of a previous question  which was marked as a duplicate. Some nice folks have referred me to solutions to similar problems. I still have a couple of questions, since one of the solutions used the Poisson kernel, which is not the desired solution in the text, because the Poisson kernel has not been introduced yet. The other solutions were for problems that were similar, but whose solutions were only partially applicable, as I understand them.  I'm working on a problem from p. 166 of Lars Ahlfors' Complex Analysis:     If $u$ is harmonic and bounded in $0 &lt; |z| &lt; \rho$, show that the origin is a removable singularity in the sense that $u$ becomes harmonic in $|z|&lt;\rho$ when $u(0)$ is properly defined.   My Solution So Far:   Here's what I've got for a non-Poisson kernel solution: We want to find a harmonic conjugate for $u$ in the punctured disk, and then we can apply big Picard to show that $0$ is a removable singularity. The problem is that the punctured disc is not simply connected.   If we take $\sigma = (u \circ \exp)(z): L \to \mathbb{C}$, where $L$ is the left half-plane, now we have $\sigma$ harmonic, and defined in a simply connected set $L$, so it has a harmonic conjugate $\tau$. Now we wish to take $v = (\tau \circ \log):\{0 &lt; |z| &lt; \rho\}\to \mathbb{C}$, but in order to do so we need $\tau$ to be $2\pi i$-periodic. This is where my proof breaks down.  One of the similar problems which readers referred me to used the following strategy: note that since $\sigma$ is $2\pi i$-periodic, and $\sigma, \tau$ satisfy the Cauchy-Riemann equations, we have that $\nabla\tau(z) = \nabla\tau(z + 2 \pi i)$ for all $z\in L$, or in other words, $\tau(z) - \tau(z+2\pi i)$ is a constant function, say, equal to $\alpha \in \mathbb{R}$. Then, the solution suggests, we can set $$g(z) = u + iv - \frac{c}{2\pi}z,$$  in order to make $g$ $2\pi i$-periodic, and then we can use $(g\circ \log)$ on $\{0 &lt; |z| &lt; \rho\}$. But the reason why this won't work for my purposes is: now we have changed $u$! On the other hand, we can't simply subtract $\frac{c}{2\pi}\Im{(z)}$, and try to leave $u$ untouched, because now $g$ is not analytic. How can we get around this?  Another strategy entirely might be to try to use the mean-value property. Ahlfors discusses that a lot in the preceding section, and his main example is the punctured disc. Since $u$ is harmonic, we know that the average over circles centered at zero is a constant... How to use exclamation marks in Minecraft commands? "It was my op" what does it mean? I heard this in  movie. It was my op. What does it mean? Googling gave me : 'Original Poster'. But that doesn't seem very meaningful. How do RAM and processor speed affect overall performance on Android? Since Android can be installed on wide range of devices, I'm wondering how important RAM and the CPU speed are to the performance of a device.  There are always tradeoffs when something is increased and another one decreased!  Is more RAM or a faster processor better for general browsing and less mulititasking?  What if I'm interested in gaming or doing lots of things at once?  Will more RAM make up for a slower processor or vice-versa? I want to show that  $$\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)} = \frac{\pi}{2\sqrt{a(a+1)}}$$  for $a &gt; 0$.  I try several methods   Substitutions to rationalize The famous $U = \tan(\frac{\theta}{2})$ Multiplying by conjugates   and other calculus techniques and still, I can not prove the equality.   Anyone can give me a hint of how should I begin to work this integral?  I'm very ashamed that I can not solve this problem. gnome-terminal fails to start (timed out) I just upgraded from Ubuntu 16.04 to 18.04. My workflow is to connect from Windows using VcXsrv to Ubuntu (ssh with X tunneling), starting a xterm on Ubuntu. Inside the xterm I am typing gnome-terminal to get a gnome-terminal. Since the upgrade this is not working anymore. I am getting $ gnome-terminal   # Couldn't register with accessibility bus: Did not receive a reply. Possible causes include: the remote application did not send a reply, the message bus security policy blocked the reply, the reply timeout expired, or the network connection was broken.   # Error constructing proxy for org.gnome.Terminal:/org/gnome/Terminal/Factory0: Fehler beim Aufruf von StartServiceByName für org.gnome.Terminal: Zeitüberschreitung wurde erreicht    or when starting with English locale $ LANG=en_US.utf8 gnome-terminal   # Couldn't register with accessibility bus: Did not receive a reply. Possible causes include: the remote application did not send a reply, the message bus security policy blocked the reply, the reply timeout expired, or the network connection was broken.   # Error constructing proxy for org.gnome.Terminal:/org/gnome/Terminal/Factory0: Error calling StartServiceByName for org.gnome.Terminal: Timeout was reached  I am not always getting the first line, but I am always gettint the timeout error. Digging more into the issue I found out  sudo gnome-terminal works gnome-terminal works when run in a xterm when locally logged in on the PC when I am also logged in locally on the PC then gnome-terminal in the remote session's xterm opens a gnome-terminal on the local session on the PC (still getting the first error line about registering with accessibility bus) when I am also logged in over rdp (Windows network connection) then a gnome-terminal in the ssh session opens a gnome-terminal in the xrdp session when I login over rdp (Windows network connection), logoff there, then gnome-terminal can open a gnome-terminal  What can I do to get gnome-terminal in a remote session's xterm to work? Can't we compute $\lim_{n \to \infty}\frac{1+ \cdots +n}{n^2}=\lim_{n\to\infty}\frac{1}{n^2}+\cdots +\lim_{n\to\infty}\frac{n}{n^2}=0$? I just learned the definition of limits, and I learned that if $\{a_n\}, \{b_n\} $ converges, then $$\lim_{n\to \infty} (a_n+b_n)=\lim_{n \to \infty} a_n+\lim_{n \to \infty}b_n$$ holds.  And my teacher said that $\lim_{n \to \infty}\frac{1+2+3+ \cdots +n}{n^2}=\lim_{n\to \infty}\frac{\frac{n(n+1)}{2}}{n^2}=\frac{1}{2}$.  But can't we compute like  $$\lim_{n \to \infty}\frac{1+2+3+ \cdots +n}{n^2}=\lim_{n\to\infty}\frac{1}{n^2}+\lim_{n\to\infty}\frac{2}{n^2}+\lim_{n\to\infty}\frac{3}{n^2}+\cdots +\lim_{n\to\infty}\frac{n}{n^2}=0+0+\cdots+0=0$$? Term for group of guys attempting to show who is alpha? Similar to pissing contest - but not in an aggressive manner. Just each guy attempting to out talk one another to show they are the leader / in charge.   Sample sentence:     Well, that meeting was more of a X than being productive with Greg trying to show he was in control. Noun-adjective reversal - was it ever in use in plain speech? Some people say that mass increases with speed, some people say that the mass of an object is independent of its speed.  I understand how some (though not many) things in physics are a matter of interpretation based on one's definitions. But I can't get my head around how both can be 'true' is any sense of the word.  Either mass increases or it doesn't, right?  Can't we just measure it, and find out which 'interpretation' is right? e.g.: (in some sophisticated way) heat up some particles in a box and measure their weight?  ------------------------------  UPDATE  Right, so I've got two identical containers, each with identical amounts of water, each on identical weighing scales, and each in the same g field. If one container has hotter water, will the reading on its scale be larger than the other? If the answer is yes, and g is constant, does this mean that the m in w=mg has got bigger? Converse of Schur's Lemma in finite dimensional vector spaces I am trying to prove (or disprove) the converse of Schur's Lemma in finite dimensional vector spaces. I am not sure if it holds in this case, but I have tried to apply the idea that proves it in representation theory (see for example Theorem 4.3  that uses Maschke's theorem or in questions  and ).  The converse of Schur's Lemma in finite dimensional vector spaces is:     Let $V$ be a finite dimensional vector space over the complex numbers. Let $S$ be a set of endomorphisms of $V$ and assume that every endomorphism $A$ of $V$ such that $$AB=BA\text{ for all }B\in S$$ is of the form $\lambda I, \ \lambda\in\mathbb{C}$. Then $V$ is a simple $S$-space.   (*an endomorphism of $V$ is a linear operator from $V$ to $V$  *$V$ is a simple $S$-space if the only $S$-invariant subspaces of $V$ are $V$ itself and the zero subspace)  What I've tried so far:  Assume to the contrary that $V$ is not a simple $S-$space. Then, there exists a subspace $W$ of $V$ such that $W\not =\{0\}$, $W\not= V$ and $W$ is $S$-invariant. Let $W'=V\setminus W$. Then, since $V$ is finite, we can easily prove that $V=W\oplus W'$ (this is my attempt to translate Maschke's theorem in vector spaces). Consequently, for every $v\in V$, there exists unique $w\in W$ and $w'\in W'$, such that $v=w+w'$. Define the projection $P:V\rightarrow V$ by $Pv=w$ for every $v\in V$.   What is left to prove is that $PB=BP$ for all $B\in S$. Then $P$ is clearly not a scalar and thus we have the contradiction we are looking for.  There is a difficulty in showing that $PB=BP$ for all $B\in S$, because $W'$ might be $S$-invariant or not. More precisely:   Let $v\in V=W\oplus W'$ and $v\not =0_V$.  If $v\in W$ then $Bv\in W$ because $W$ is $S$-invariant, therefore: $PBv=Bv$ and $BPv=Bv$.  If $v\in W'$ then $Pv=0$, hence $BPv=B\cdot 0=0$, and  (i) if $Bv\in W'$ then $PBv=0$.  (ii) if $Bv\in W$ then $PBv=Bv$ and this is where the problem occurs.  Any hints, ideas or counterexamples would be very helpful. Using image sequence of medical scans as volume data in cycles I know it's possible to do this in BI, but I need your help to do this in cycles. For now, I have a setup to use an image sequence as volume but it only shows 1 slice/frame. I'm using generated texture coordinates for easy mapping (0-1 on each axis). The images are mapped on XY and I would like to use normalized frame# as Z coordinate to map the sequence as a volume.       Images sequence can be found  (I'm using 16-bit TIF files of MRBrain in this exemple).    Any solution is welcome but try to keep it simple and user friendly :).  Final Edit: I had a hard time choosing between Lemon's and PGmath's answers. Lemon's is more user friendly and flexible, and uses the image sequence without any change inside blender. PGmath's uses a 3rd party converter to make the sequence 1 single image, but allows GPU rendering. Well, ImageMagick's command line is : "convert +append *.tif anim.tif" from the folder where the sequence is unzipped. Not so hard. That's why I accepted PGmath's answer.      Now, i played with it, and here's the result.  The "volume" object contains the volume shader (playing with the color ramp gives funny things) and the boolean modifiers (turn them off to see the complete volume, but reduce the final density multiplier for better view). The "Slice Control" arrow empty controls the slice position/rotation. The "Texture Mapping" object controls...the mapping, if it need some tunning (position, rotation, scale, and then tiling if necessary?). How do I copy to the clipboard in JavaScript? What is the best way to copy text to the clipboard (multi-browser)? I have tried: function copyToClipboard(text) {     if (window.clipboardData) { // Internet Explorer         window.clipboardData.setData(&quot;Text&quot;, text);     } else {         unsafeWindow.netscape.security.PrivilegeManager.enablePrivilege(&quot;UniversalXPConnect&quot;);         const clipboardHelper = Components.classes[&quot;@mozilla.org/widget/clipboardhelper;1&quot;].getService(Components.interfaces.nsIClipboardHelper);         clipboardHelper.copyString(text);     } }  But in Internet Explorer it gives a syntax error. In Firefox, it says unsafeWindow is not defined. A nice trick without using : Where are my installed applications? I've seen sign in methods using  as an authentication provider, for example : (AWS Cloud9)        Is it possible to authenticate into Stack Exchange using this method? At the moment there's other methods such as OpenID, Google, Yahoo and Facebook, but it would be interesting to also have the GitHub authentication. Is there a $(6,9,3)_2$ code? Does there exist a $1$-error-correcting binary code with block length $6$ and $9$ codewords?   The Hamming bound says that for any code $C$ with those parameters, $|C| \le \frac{2^6}{1+6} \approx 9.14$. So, we can't rule out the existence of such a code using the Hamming bound. The Singleton bound says that for any code $C$ with those parameters, $|C| \le 2^{6-3+1} = 16$, so we can't rule out the existence of such a code using the Singleton bound. Also, this can't be a linear code, since $\log_2 9$ isn't an integer. That rules out the possibility of just enumerating the possible generator matrices.   I feel a bit silly asking this, but the direct approach (assume $000\ 000$ is in the code, then all others must have Hamming weight $\ge 3$, ...) becomes unmanageable quickly. How else can I go about this? Deriving the joint probability density function from a given marginal density function and conditional density function We have a multivariate normal vector $\mathbf{X} \sim \mathcal{N}(\boldsymbol{\mu}, \boldsymbol{\Sigma})$, where $\mathbf{X} =  \left[ \begin{matrix} X_1 \\ X_2 \\ \dot\\\\ \dot\\\\ X_n \end{matrix} \right]$,    $\boldsymbol{\mu} =  \left[ \begin{matrix} \mu_1 \\ \mu_2 \\ \dot\\\\ \dot\\\\ \mu_n \end{matrix} \right]$, and  $\boldsymbol{\Sigma} =  \begin{bmatrix} \sigma_1^2 &amp; \sigma_1\sigma_2 &amp; \dots &amp; \sigma_1\sigma_n   \\ \sigma_2\sigma_1 &amp; \sigma_2^2 &amp; \dots &amp; \sigma_2\sigma_n  \\ \dot\\\\ \dot\\\\ \sigma_n\sigma_1 &amp; \sigma_n\sigma_2  &amp; \dots &amp; \sigma_n^2 \end{bmatrix} $ is the covariance matrix.  Y is a random scalar of the form $Y = \beta_0 + \boldsymbol{\beta^TX} + \epsilon$, where $\epsilon \sim \mathcal{N}(0, \sigma_\epsilon^2)$  How can we go about deriving the joint probability density function(pdf) of the random vector $(Y, \boldsymbol{X^T})$?  The marginal multivariate pdf of $\boldsymbol{X}$ is given by  $ pdf(\boldsymbol{x}) = \frac{1}{\sqrt{(2\pi})^n. \begin{vmatrix} \boldsymbol{\Sigma} \end{vmatrix}}.exp\left[ \frac{-1}{2}\boldsymbol{(x-\mu)}^T \boldsymbol{\Sigma^{-1}}\boldsymbol{(x-\mu)} \right]$  The conditional probability density function of $Y \,\mid \, \boldsymbol{X=x}$ is given by  $pdf(y\mid \boldsymbol{X=x}) = \frac{1}{2\pi\sigma_\epsilon}.exp\left[  \frac{-1}{2\sigma_\epsilon^2}(y - \beta_0 - \boldsymbol{\beta^Tx})^2  \right]   $  To compute the joint probability density function, we can use the fact that   $pdf(y, \boldsymbol{x})=  pdf(y\mid \boldsymbol{X=x}).pdf(\boldsymbol{x})$  I am unable to do the witty manipulation required in the algebra to arrive at the result the joint distribution will have a probability density function of a normal distribution. All questions previously asked seem to address the problem of deriving the conditional density function given the joint probability density function, for example,   Edit:  Multiplying the pdfs, we get $pdf(y, \boldsymbol{x}) = \frac{1}{\sqrt{(2\pi})^n. \begin{vmatrix} \boldsymbol{\Sigma} \end{vmatrix}}. \frac{1}{2\pi\sigma_\epsilon}.exp\left[ \frac{-1}{2}\boldsymbol{(x-\mu)}^T \boldsymbol{\Sigma^{-1}}\boldsymbol{(x-\mu)} + \frac{-1}{2\sigma_\epsilon^2}(y - \beta_0 - \boldsymbol{\beta^Tx})^2 \right]$   Like the comment below mentions, most manipulations dealing with normal pdfs involve completing squares; but since this specific algebraic expression contains $\boldsymbol{x}$ in both terms within the exp term, I'm unsure on how to go about completing the squares. About the number format in ticks