I recently found out about , which appears to be a normal beta Stack Exchange site (I was able to sign up immediately and receive the standard Association Bonus, etc.), but I noticed that it neither appears on the  list nor the Communities list that appears when you click the SE hamburger menu logo.  At first, I assumed that CS50 was probably a private beta, but there is no corresponding proposal on Area 51 for anything called CS50. The site also seems to have questions dating back to at least 2014, which is a very long time indeed for a site to stay in private beta.  What publicly accessible Stack Exchange sites currently exist in the Stack Exchange network, but do not appear in the standard listings? That is, these are sites that anyone can, in theory, sign up for, but which are not normally disclosed to users via the standard methods for informing users about available sites. This could be because they are highly niche, controversial, or have been explicitly removed from the listings for some other reason.
I wanted a list of sub-domains of Stack Exchange that are not mentioned over here   Some of the sub-domains I found which were not mentioned are       Are there any other which are not mentioned?    Mentions by , ,             Other mentions below which are not mentioned in "All Sites" but are mentioned in the drop down menu on the top left of the Stack Exchange website         Any other website that's left out ?
Fourier transform of $\exp(-t^2)$ using contour integration. General identity for a double summation theorem I've been reading a research paper, and I'm interested in generalizing a certain theorem but I can't seem to understand how the following results are derived:  $$\sum_{i=1}^{n}\sum_{j\neq i}^n \frac{1}{z_i-z_j}=0$$  $$\sum_{i=1}^{n}\sum_{j\neq i}^n \frac{z_i}{z_i-z_j}=\frac{1}{2}n(n-1)$$ $$\sum_{i=1}^{n}\sum_{j\neq i}^n \frac{z_i^2}{z_i-z_j}=(n-1)\sum_{i=1}^n z_i$$ $$\sum_{i=1}^{n}\sum_{j\neq i}^n \frac{z_i^3}{z_i-z_j}=(n-1)\sum_{i=1}^n z_i^2 + \sum_{i&lt;j}^n z_i z_j$$ Given that $a$~$\mathcal{N}(\mu, \Sigma)$ and $\mu$~$\mathcal{N}(0, \sigma^2 I)$, Is $p(a|\Sigma, \sigma)=\int p(a|\mu, \Sigma)p(\mu|\sigma)d\mu$ again a Normal distribution? And how to estimate the mean and variance respectively? Thanks a lot! Suppose I sold an option on stock X, with strike price Y, and expiry date Z. Suppose I later buy an option on the same stock, with the same strike price, and the same expiry date. Do these two positions cancel out each other (i.e. both contracts cease to exist anywhere), or will I end up with two separate contracts?  Is it the same with the reverse situation? e.g. I bought an option on stock A, with strike price B, and expiry date C. If I later sell an option on the same stock, with the same strike price and expiry date, will I end up with two contracts (one long position, one short position)? Story about a trip to a Valley of Sound I read a short story, can’t remember if it was part of a larger narrative or just on its own.  It was about a crew that was taking a bunch of space tourists to a planet that had plants with seeds in glass-like structures that make certain sounds. The main character in this story feels connected to this planet because there’s a Valley of Sound, where there’s a large number of these glass plants that make all sounds imaginable.  People can almost pick out a loved one’s voice for instance. I absolutely love this concept but can’t for the life of me remember what the story/book was called! Set an alias on Bourne shell I am using a scrtipt with instruction 'SH -c command', where command is an alias from bash, this alias can not be interpreted into Bourne shell. ¿Do you have any idea for set this alias into Bourne shell and getting that it works in the new shell? Thanks everyone ! How can I clip an image via a bezier-path? Integrating trigonometric function problem $\int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx$ \begin{eqnarray*} \int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx &amp;=&amp;  \int \frac{(3\sin x+2\cos x)/\cos x}{(2\sin x+3\cos x)/\cos x}dx\\ \\ &amp;=&amp; \int \frac{3\tan x +2}{2\tan x +3} dx\\ &amp;&amp; u = \tan x \text{ and } du = \sec^2 x \ dx  \end{eqnarray*} Am I going in the right direction with this one? It seems like not. Stack Exchange Android app keeps replaying notifications Version of the app: 1.0.51. (It is up to date at the time of writing according to Google Play.)  This started happening after the network problem that happened earlier this morning. Both my tablet and my phone are popping up notifications for events (comments, and probably edits too) that happened prior to the network issue. I had received notifications for these events already and had already cleared them before the network issue happened.  The repeated notifications do not show up in the bar displayed at the top of the Stack Exchange sites. In other words, I get notified again in Android but do not get notified again on the SE web sites. How can I make Ubuntu 18.04 / 18.10 desktop use Unity (be like Ubuntu 14.04)? Chirp Transform and Convolution I was reading about the discrete fourier transform from the CLR algorithms book and I came upon an exercise whose hint confuses me. The exercise reads as follows:  The chirp transform of a vector $a=(a_0,a_1,\ldots,a_n)$ is a vector $y=(y_0,y_1,\ldots,y_n)$, where $y_k=\sum_{j=0}^{n-1}a_j (z^{k})^j$ and $z$ is any complex number. The DFT is therefore a special case of the chirp transform, obtained by taking $z=\omega_n$. Prove that the chirp transform can be evaluated in time $O(n\lg n)$ for any complex number $z$. (Hint: use the equation $$y_k = z^{k^2/2}\sum_{j=0}^{n-1}(a_j z^{j^2/2})(z^{-(k-j)^2/2})$$ to view the chirp transform as a convolution.  My question is about what is a convolution. I am familiar with taking two vectors $a=(a_0,a_1,\ldots,a_n)$ and $b=(b_0,b_1,\ldots,b_n)$ and forming the convolution $c=(c_0,c_1,\ldots,c_n)$ where $c_k = \sum_{j=0}^k a_j b_{k-j}$. Here the index of $j$ runs from $0$ to $k$ unlike in the hint given above where the index runs from $0$ to $n-1$. In what sense is the hint above a convolution? Bridged adapter in Virtual Box - cannot obtain IP I have virtual box with RHEL installed. My virtual box is running on windows 7 64 -bit. I configured bridged connection, but when I am restarting network interfaces, I receive following error:  Determining IP information for eth1... failed   On Windows XP it works without problems. I disabled Windows Firewall and installed latest Virtual Box, still without success.   NAT is working without any problems. What did I missed?  UPDATE  Ouput of ifconfig -a is:  eth0      Link encap:Ethernet  HWaddr 08:00:27:B6:FC:69           inet addr:10.0.2.15  Bcast:10.0.2.255  Mask:255.255.255.0           inet6 addr: fe80::a00:27ff:feb6:fc69/64 Scope:Link           UP BROADCAST RUNNING MULTICAST  MTU:1500  Metric:1           RX packets:430 errors:0 dropped:0 overruns:0 frame:0           TX packets:346 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:1000           RX bytes:36520 (35.6 KiB)  TX bytes:31455 (30.7 KiB)  eth1      Link encap:Ethernet  HWaddr 08:00:27:C3:84:4A           inet6 addr: fe80::a00:27ff:fec3:844a/64 Scope:Link           UP BROADCAST RUNNING MULTICAST  MTU:1500  Metric:1           RX packets:1464 errors:0 dropped:0 overruns:0 frame:0           TX packets:17 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:1000           RX bytes:135419 (132.2 KiB)  TX bytes:4230 (4.1 KiB)  lo        Link encap:Local Loopback           inet addr:127.0.0.1  Mask:255.0.0.0           inet6 addr: ::1/128 Scope:Host           UP LOOPBACK RUNNING  MTU:16436  Metric:1           RX packets:0 errors:0 dropped:0 overruns:0 frame:0           TX packets:0 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:0           RX bytes:0 (0.0 b)  TX bytes:0 (0.0 b)   Output of route -n  Kernel IP routing table Destination     Gateway         Genmask         Flags Metric Ref    Use Iface 10.0.2.0        0.0.0.0         255.255.255.0   U     0      0        0 eth0 169.254.0.0     0.0.0.0         255.255.0.0     U     1002   0        0 eth0 0.0.0.0         10.0.2.2        0.0.0.0         UG    0      0        0 eth0   Output of dhclient -v eth1 is:  Listening on LPF/eth1/08:00:27:c3:84:4a Sending on   LPF/eth1/08:00:27:c3:84:4a Sending on   Socket/fallback DHCPDISCOVER on eth1 to 255.255.255.255 port 67 interval 6 (xid=0x6afb6c4c) DHCPDISCOVER on eth1 to 255.255.255.255 port 67 interval 9 (xid=0x6afb6c4c) DHCPDISCOVER on eth1 to 255.255.255.255 port 67 interval 15 (xid=0x6afb6c4c) DHCPDISCOVER on eth1 to 255.255.255.255 port 67 interval 18 (xid=0x6afb6c4c) DHCPDISCOVER on eth1 to 255.255.255.255 port 67 interval 13 (xid=0x6afb6c4c) No DHCPOFFERS received. No working leases in persistent database - sleeping.   Moreover the same machine with the same configuration works well on Windows XP. The problem is only on Windows 7. The function $f(z)=z+\frac{1}{z}$ maps $\Omega$ biholomorphically on the upper plane The question states:     Let $\Omega$ be the upper half plane minus the unit disk. Show that the map $f: z \mapsto z+1/z$ maps $\Omega$ biholomorphically on the upper plane. Describe the preimages of the lines $\mathrm{Im}(z)=\text{a constant}&gt;0$.   So how exactly does one show the mapping is biholomorphic. I'm not sure where I should start, so unfortunately I haven't made any progress Can I use a custom Splash screen? The default Splash screens are cool, but I'd like to add some of my own work to remind me of how awesome I am (no, not really).    Is it possible to customize/change the splash screen? If $f:\mathbb R \to \mathbb R$ satisfies $f'(x) <L$ for some $L<1$, then $f$ has a unique fixed point Assume that $f: \mathbb{R} \rightarrow \mathbb{R}$  is differentiable.  If there exists an $L&lt;1$ such that for each $x \in \mathbb{R}$, $f'(x)&lt;L$, prove that there exists a unique point $x$ such that $f(x)=x$.  So far I have that where I fixed $a$ to be zero and came up with three cases to show existence for  $f(0) = 0$, $f(0) &lt; 0$, and $f(0) &gt; 0$  $f(0)=0$ was fairly trivial following from the mean value theorem but the other two cases I was only able to eliminate half of the possible fixed points by contradiction and I need help showing it explicitly.  Edit as to avoid confusion here is the work I have garnished so far:  From the mean value theorem it follows for some arbitrary $a,b \in \mathbb{R}$ and $a\neq b$ that there exists a point $c$ such that $a&lt;c&lt;b$ or $b&lt;c&lt;a$ then implies that $\frac{f(b)-f(a)}{b-a}=f'(c)$ and this is because it is differentiable on $\mathbb{R}$ and, by convention, is also continuous on $\mathbb{R}$ as well.  Then the result of $\dfrac{f(b)-f(a)}{b-a}=f'(c)$ then follows for $L&lt;1$ such that $\forall x \in \mathbb{R}$ means $f'(x)&lt;1$ from our hypothesis. Which then follows  (*) $\dfrac{f(b)-f(a)}{b-a}=f'(c)&lt;L&lt;1 \Rightarrow f(b)-f(a)&lt;L(b-a)&lt;b-a$  Now without loss of generality choose $a=0$ and let $b$ be some arbitrary value in $\mathbb{R}$ then there are three cases that need to be considered for the value of $f(a)$:  (i) $f(0) = 0$, (ii) $f(0) &lt; 0$, (iii) $f(0) &gt; 0$  (i) For $f(0)=0$   Then $0$ is a fixed point of point of $f(x)$ and it follows from (*) that if $f(0)=a=0$ then $f(b)-f(0)=f(b)&lt;Lb&lt;b$ which shows that $f(b)&lt;b$ thus shows that for any arbitrary value $b\in \mathbb{R}$ that no $f(b)=b$ thus shows that $0$ is the only fixed point if $f(0)=0$  (ii) For $f(0)&lt;0$  Suppose there exists a fixed point $d&gt;0$ then:  $\frac{f(d)-f(0)}{d}&lt;L&lt;1 \Rightarrow d-f(0)&lt; dL&lt;d \Rightarrow -f(0)&lt;d(L-1)&lt;0\Rightarrow 0&lt;-d(L-1)&lt;f(0)$ but this contradicts that $f(0)&lt;0$  thus no other fixed point exists if $d&gt;0$ when $f(0)&lt;0$  (iii) For $f(0)&gt;0$ What is the difference between Unix, Linux, BSD and GNU? What is the very fundamental difference between Unix, Linux, BSD and GNU?  Unix was the earliest OS, so the term 'Unix like' is understandable, since they have kernel, file system structure, most of the commands, users etc are same as Unix.  Still why are they different? What made them set apart? Is it the kernel? Rank of the $n \times n$ matrix with ones on the main diagonal and $a$ off the main diagonal I want to find the rank of this $n\times n$ matrix \begin{pmatrix} 1 &amp; a &amp; a &amp; \cdots &amp; \cdots &amp; a \\ a &amp; 1 &amp; a &amp; \cdots &amp; \cdots &amp; a \\ a &amp; a &amp; 1 &amp; a &amp; \cdots &amp; a \\ \vdots &amp; \vdots &amp; a&amp; \ddots &amp; &amp; \vdots\\ \vdots &amp; \vdots &amp; \vdots &amp; &amp; \ddots &amp; \vdots \\ a &amp; a &amp; a &amp; \cdots  &amp;\cdots &amp; 1 \end{pmatrix}  that is, the matrix whose diagonals are $1's$ and $a$ otherwise, where $a$ is any real number.  My first observation is when $a=0$ the rank is $n$ and when $a=1$ the rank is $1.$ Then I can assume $a\neq 0, 1$ and proceed row reduction to find its pivot rows. I obtain  \begin{pmatrix} 1 &amp; a &amp; a &amp; \cdots  &amp; a \\ 0 &amp; 1+a &amp; a &amp; \cdots &amp; a \\ 0 &amp; a &amp; 1+a &amp; \cdots &amp; a \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 0 &amp; a &amp; a &amp; \cdots  &amp; 1+a \end{pmatrix} by subtracting the first row multiplied $a$ for each row below the first, and then divides the factor $(1-a)$,  and stuck there. Any hints/helps? Find the value of $g'(0)$. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function with $f(1) = 5$ and $f(3) = 11$. If $g(x) = \int_1^3 f(x+t)dt$ then find $g'(0).$   Please explain your answer. I read a novel when I was 12 or 13 (late 80's, early 90's) about two timelines combining on Earth.  It was a full length novel for adults that I swiped from my mom (she doesn't remember it at all.)  I've been trying to figure out what the title is for some time but my google-fu is unable to find the right search terms.    The book was published in the 80's or perhaps early 90's.  The story: It was set on Earth and featured two timelines running concurrently (in the 80's?)  Timeline 1:  The nuclear missile crisis had gone badly and a low level nuclear war had more or less destroyed the US, but small towns still survived but are slowly fading away.  Much of the story was about the scarcity of goods, and the protagonist in the main thread of the book started the story by finding a 6 pack of Coke buried in a ruined house.  Throughout the story people were finding things that had grown scarce (deer, bullets, human beings.)  One strange scene I recall involved mutant/radioactive sweet potatoes that had driven a man insane.  Eventually people in this timeline began having dreams of their lives in Timeline 2.  Some of the people were not having dreams, and it was eventually discovered that they had died in Timeline 2.    Timeline 2:  There is panic in a world recognizable as our 1980's as nuclear war draws near.  Since more and stronger missiles have been developed, the end of humanity is at hand.  Things start disappearing from that world as the war begins.  Timeline 2 was mostly seen as vignettes of panicked people running around.  Combination of Timelines: In a weird, incongruous, almost surely poorly written scene, Gaia (yes, Earth itself) is seen to be combining a dying timeline with a doomed timeline in order to save humanity.  At the end of the novel, kids from Timeline 2 are transported into Timeline 1 in order to repopulate.  The new kids drink all the Coke, in case you were wondering...  Any help finding the title would be welcome.  I recall enjoying the book a lot when I was a kid, but I suspect my mom traded it in at our local used book store because at some point it just disappeared. I was wondering why binary crossentropy can be used as the loss function in autoencoders trained on (normalized) images, e.g.  or ? I know that binary crossentropy can be used in binray classification problems where the ground-truth labels (i.e. $y$) are either 0 or 1 and therefore when predictions (i.e. $p$) are correct, in both cases, the loss value would be zero:  $$ BCE(y,p) = -y.log(p) - (1-y).\log{(1-p)} $$  $$ BCE(0,0) = 0, BCE(1,1) = 0 $$  However, binary crossentropy does not have a value of zero when neither of its arguments are both zero or one, which is the case for an autoencoder with ground-truth labels in range $[0,1]$ (i.e. assuming the input data has been normalized in this range). I thought a regression loss function such as mean squared error or mean absolute error must be used instead, which have a value of zero when labels and predictions are the same. What am I missing here?