Why doesn't Stack Overflow have a "Chat" link like the rest of the sites?
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"method driver /usr/lib/apt/methods/https could not be found" update error sudo apt-get update E: The method driver /usr/lib/apt/methods/https could not be found. michael@Ubuntu:~$ sudo apt-get clean michael@Ubuntu:~$ cd /var/lib/apt michael@Ubuntu:/var/lib/apt$ sudo mv lists lists.old michael@Ubuntu:/var/lib/apt$ sudo mkdir -p lists/partial michael@Ubuntu:/var/lib/apt$ sudo apt-get clean michael@Ubuntu:/var/lib/apt$ sudo apt-get update E: The method driver /usr/lib/apt/methods/https could not be found. E: Could not get lock /var/lib/dpkg/lock - open (11: Resource temporarily unavailable) E: Unable to lock the administration directory (/var/lib/dpkg/), is another process using it? michael@Ubuntu:/var/lib/apt$ sudo apt-get update E: The method driver /usr/lib/apt/methods/https could not be found.   Why am I getting these errors when doing an apt-get update? Word for someone who repeats words This post has been locked; locked posts can't be voted on Study the convergence of the following series: $$\sum_{n=2}^\infty \frac{1}{n^\alpha \cdot\ln^\beta(n)} \text{ where }\alpha,\beta \geq 0 $$   Applying d'Alembert criterion I have that $$ \lim_{n\to\infty} \frac{n^\alpha  \ln^\beta(n)}{(n+1)^\alpha  \ln^\beta (n+1)} = \lim_{n\to\infty} \left(\frac{n}{n+1}\right)^\alpha\left(\frac{\ln(n)}{\ln (n+1)}\right)^\beta = 1$$ so the nature of the series is inconclusive.  If $\alpha = \beta = 0$, then the series diverges, since $\sum_{n=2}^\infty 1 = \infty$.  Should I study the rest of the cases (i.e. if $\alpha = 0, \beta &gt; 0$ the root test and the ratio test are also inconclusive). What is the best form to study the series?.  Thanks in advance When playing with the complete elliptic integral of the first kind and its Fourier-Legendre expansion, I discovered that a consequence of $\sum_{n\geq 0}\binom{2n}{n}^2\frac{1}{16^n(4n+1)}=\frac{1}{16\pi^2}\,\Gamma\left(\frac{1}{4}\right)^4 $ is:     $$\int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x^2)}}\,dx = \tfrac{1}{32}\sqrt{2\pi}\,\Gamma\left(\tfrac{1}{4}\right)^2\tag{A}$$   which might be regarded as a sort of Ahmed's integral under steroids.  I already have a proof of this statement (through Fourier-Legendre expansions), but I would be happy to see a more direct and elementary proof of it, also because it might have some consequences about the moments of $K(x)$ of the form $\int_{0}^{1}K(x)\,x^{m\pm 1/4}\,dx$, which are associated with peculiar hypergeometric functions. Matrices that can only be permutation matrices How to prove that if $A$ and $B$ are two $n×n$ matrices with non negative integer entries such that $AB=I$ then $A$ and $B$ are permutation matrices?  Permutation matrices are matrices with the columns just a permutation of the identity matrix. How do I proceed. Can somebody give hint? A while ago, I had a discussion with a colleague about how to insert values in STL . I preferred map[key] = value; because it feels natural and is clear to read whereas he preferred map.insert(std::make_pair(key, value)). I just asked him and neither of us can remember the reason why insert is better, but I am sure it was not just a style preference rather there was a technical reason such as efficiency. The  simply says: &quot;Strictly speaking, this member function is unnecessary: it exists only for convenience.&quot; Can anybody tell me that reason, or am I just dreaming that there is one? I've tried several things already,  std::stringstream m; m.empty(); m.clear();   both of which don't work. I have two raster images, Boolean classified (1 or 2).  They overlap in given areas.   I need a raster that is only the overlapping sections.  Using Arcinfo 10.  Answer can only use raster methods, no converting to polygons, points, etc... $g^k$ is a primitive element modulo $m$ iff $\gcd (k,\varphi(m))=1$ Already I have lost all hope of looking at late answers etc. It's just a parade of meh. I can't make any kind of intelligent decision because I can't see the other answers to the question, for instance. There's no Looks Good, only Skip, and those feel different to me. So I've been more interested in .  But people keep approving crap while I'm looking at it. Two or three "the post was already approved; please visit to edit" and I'm just "screw this whole thing".  Can we do something like one reject bounces all existing votes: you need three approves in a row? Because a lot of people are stupidly approving and actually driving me away from the queue.    Note this is less about letting bad edits through and more about keeping reviewers motivated. What is the definition of layer in neural network? Consider a complete metric space $(X,d)$ and $T\colon X\to X$. Suppose there exists $n\in\mathbb{N}$ such that the n-th power of $T$ is $q$-contractive. Show that then $T$ has exactly one fixed point $\overline{x}\in X$.      The n-th power is defined inductively: $$ T^{n+1}(x):=T(T^{n}(x)), n\in\mathbb{N}. $$ And for a function $T\colon X\to X$, $q$-contractive means  $$ \exists 0\leq q&lt;1~\forall~x,y\in X: d(T(x),T(y))\leq q d(x,y). $$    Now to the proof.  I think I have to show, that $T$ is q-contractive, because then it follows with Banach, that $T$ does have exactly one fixed point. So I have to show, that there exists a $0\leq q&lt;1$ so that for all $x,y\in X$ it is $$ d(T(x),T(y))\leq q\cdot d(x,y). $$ And most likely I have to use the q-contractivity of $T^n$, i.e. that there exists a $0\leq q &lt;1$ so that for all $x,y\in X$ it is $$ d(T^n(x),T^n(y))\leq q\cdot d(x,y). $$  Can you help me? Area of the Limiting Polygon Start with an equilateral triangle with unit area. Trisect each of the sides and then cut-off the corners. In this case, we get a regular hexagon - see the picture below. Next, trisect each of the sides of the hexagon and cut-off the corners. This will give a dodecagon, but not a regular one.   Continue this process ad infinitum.  What is the area of the limiting "polygon"?  My first thought was that it would be the circle tangent to the midpoint of each side. However, by making some pictures, it becomes clear quite quickly that the limiting polygon is not a circle and, in fact, might perhaps be an ellipse. The problem with that is that while there is a unique circle passing through any three points in general position, I need five generic points to specify a conic.   Can anyone suggest any hints as to how to find the area? By "hint" I don't mean "what you think might work", I mean "what you have tried and you know will work". Thanks in advance. Birational map from elliptic curve keeps smoothness? Let $E$ be a elliptic curve that is , genus $1$ smooth curve with base point. And suppose that the map $φ$, which sends $E$ to $E'$, is birational. Birational means there are two rational maps on both direction which composites identity. Then, can we say that $E'$ is smooth? Every time Windows 10 runs updates (which I see no way to control), it updates my graphics driver.  I like the old one because it is faster.  So, I have to reinstall it.  This is getting to be a pain.  Is there any way to stop Windows 10 from updating it again? A little brain fart What is a nicer, less immature saying?  I love using 'A Freudian slip', but that is only applicable in certain situations. Is my sugar daddy scamming me Sound of a next scene starts one moment earlier than video I saw this several times in different movies, but I don't understand what is this supposed to mean. The main idea is when the scene is about to change, for one second or so we still see the picture from the old scene, but the sound is already coming from from the upcoming scene. Why do they do this? What's the purpose of this technique? One-sided limits of $f'(x)$ at a point vs. one-sided limits of the difference quotient at that point I know that if a function is piecewise, then when differentiating it one needs to deal separately with the points of intersection of the pieces, because the derivative may not be defined there.  Also, if a one-sided limit of $f'(x)$ at these "crucial points" is infinite or doesn't exist, it doesn't preclude the limit of the difference quotient from existing and being finite. For example, take $f(x)=x^2\sin\frac1x, f(0)=0$: $f'(x)= 2x\sin\frac1x-\sin\frac1x $ has no limit as $x\to0$ but $\lim\limits_{h\to0}\frac{f(h)-f(0)}{h}=0=f'(0)$.  On the contrary, looking at $f(x)=\sin x +\lvert \sin x\rvert $ on $[0,2\pi]$ one has $f'(x)=2\cos x$ on $(0,\pi) $ and $f'(x)=0$ on $(\pi,2\pi)$ so $\lim\limits_{h\to0^+}f'(\pi+h)\ne\lim\limits_{h\to0^-}f'(\pi+h)$, but they are both finite. And in this case it's easy to find that they agree with the one-sided limits of the difference quotient. Why? Does the finiteness of the former imply that they agree with the latter? Is this also true if the one-sided limits of $f'$ at the crucial points agree with each other?