Using the Android app, I frequently select from the tags list to view recent questions on subjects I might be able to help with.  However, after viewing a question, tapping back (either the phone's default or the arrow in the top bar) returns me to the tags list, not the questions as one would expect. It makes for a frustrating user experience.  Any chance this could be altered in a forthcoming version?
This looks a little bit like a variation on  but specific to the tags view.  Reproduction (app version 1.0.71):   Select any site (say Meta) from the sidebar Switch to the tags view using the top bar Choose a tag, any tag ( is nice)   optionaly refine your search with more terms  Go to any question found, go back   you end on the unfiltered tags view  Go back again   lucky you, your search is here
Determine if $\sum_{n=1}^\infty \frac{n+4^n}{n+6^n}$ converges I am having trouble understanding the estimation of an AR process. In some textbooks, the AR(1) process is defined as follows: $y_{t}=\theta y_{t-1}+ϵ_t$ (which does not contain a constant). So the OLS estimator is biased. I am confused about the cause of the bias. It is explained that $y_{t-1}$ is dependent on $ϵ_{t-1}$ although it is independent of $ϵ_t$. However in linear regression, if the equation does not contain a constant, we cannot make sure the expectation of disturbance is zero. So the OLS estimator is bias without a constant. Does it mean that the OLS estimator is unbiased if I add a constant in AR process? How can I easily clone my entire system? I am running an Ubuntu 14.04 LTS and I experienced some issues with my hard drive. I bought a new HDD of the same size (1TB), and I intend to move all my data (softs, work, pics, etc.) on the new HDD.  What is the easiest way to do that? How do I insert a nice tilde that is centered in the middle in a lstlisting? I found another thread about  but I can't apply it because the method described there only works outside of lstlisting as lstlisting displays all the text literally. Any ideas?  Here is how the tilde looks in my document at the moment - it's at the very top of the font and I would like it to be in the middle (like ~ this).    Here is the LaTeX for this fragment:  \section{Remove duplicate, consecutive lines (emulate "uniq")}  \begin{lstlisting} awk 'a !~ $0; { a = $0 }' \end{lstlisting}  Variables in Awk don't need to be initialized or declared before they are being used. What does E mean in 9.0122222900391E-5? I often find this at the bottom of pages.  Page generated in 0.00013899803161621  Sometimes, I come across  Page generated in 9.0122222900391E-5  What does that time mean? I tried searching Wikipedia for E and maths but found the . My guess is E stand for Exponential and -5 is the power it is raised to. And the displayed time is a really small number. But that doesn't make sense when compared to the other time in the question. 0.00013899803161621 is bigger than 9.0122222900391E-5. If it means x times $10^{-5}$, then 9.0122222900391E-5 will be 0.000090122222900391 which is smaller than 0.00013899803161621. What does E stand for? Let $m$ be Lebesgue measure and $a \in R$. Suppose that $f : R \to R$ is integrable, and $\int_a^xf(y) \, dy = 0$ for all $x$. Then $f = 0$ a.e. This is a corollary to a proof in Bass, but I don't understand why it follows from the proof he gives. I follow everything up until the last statement. Why is it that proving that the integral is $0$ for all Borel measurable sets guarantees it is true for all Lebesgue measurable sets? Am I missing something vital about the relationship between Lebesgue measurable sets and Borel measurable sets? Statement: Let $m$ be Lebesgue measure and $a \in\mathbb R$. Suppose that $f : \mathbb R \to\mathbb R$ is integrable, and $\int_a^xf(y) \, dy = 0$ for all $x$. Then $f = 0$ a.e. Proof: We prove the conclusion first for intervals, then for unions of intervals, then for open sets, then for Borel measurable sets. For any interval $(c,d)$, we have $$\int_c^df = \int_a^d f - \int^c_a f = 0.$$ By linearity, if $G$ is the finite union of disjoint open intervals, then $\int_G f = 0$. Now, if $G$ is open, then $G$ is the countable union of disjoint, open intervals, $G = \bigcup^n_{i=1}I_i$. We have $$\int_G f = \int \lim_{n\to\infty} f \cdot \chi_{\bigcup^n_{i=1} I_i}$$ and since $f_n$ is integrable for all $n$, and $|f_n| \le f$ for all $n$, by the dominated convergence theorem, $$\int_G f = \lim_{n\to\infty} \int_{\bigcup^n_{i=1}I_i} f =0.$$ Now, if $G_n$ is a decreasing sequence of open sets converging to $H$, since $|f \cdot \chi_{G_n}| \le f \cdot \chi_{G_1}$ for all $n$, we again have by the DCT that $\int_H f = 0 = \lim_{n\to\infty} \int_{G_n}f = 0$. Finally, if $G$ is a Borel measurable set, then we know that there exists a sequence $G_n$ of decreasing, open sets, that converge to some set $H$, where $H \setminus G$ is a null set. Thus, $$\int_G f = \int_H f = 0.$$ We have found that for every Borel measurable set $G$, $\int_G f =0$. Since $f$ is real valued and measurable, $f = 0$ a.e. Geometric interpretation of the multiplication of complex numbers? I've always been taught that one way to look at complex numbers is as a Cartesian space, where the real part is the $x$ component and the imaginary part is the $y$ component.  In this sense, these complex numbers are like vectors, and they can be added geometrically like normal vectors can.  However, is there a geometric interpretation for the multiplication of two complex numbers?  I tried out two test ones, $3+i$ and $-2+3i$, which multiply to $-9+7i$.  But no geometrical significance seems to be found.  Is there a geometric significance for the multiplication of complex numbers? How to find out the terminal command of an application? I want to use some applications shown on the Unity Launcher in a terminal, but I need to know the appropriate command to run them. How can I achieve that? Why is asymmetric cryptography bad for huge data? I've been told that asymmetric cryptography requires that the message to be encrypted be smaller than its key length. Why is this?  I know about hybrid encryption, which . But I still want to know why public-key cryptography needs the data to be shorter than the key length. Ackermann function - how to calculate the number of times it calls itself Purely for my own amusement I've been playing around with the .  The Ackermann function is a non primitive recursive function defined on non-negative integers by:  $A(m,n) = n+1$, if $m=0$  $A(m,n) = A(m-1,1)$, if $m&gt;0$ and $n=0$  $A(m,n) = A(m-1,A(m,n-1))$, if $m&gt;0$ and $n&gt;0$  I'm not interested in values of the function itself, rather I'm interested in the number of recursions the function takes to arrive at a value. For example:  $A(1,2) = A(0,A(1,1)) = A(0,A(0,A(1,0))) = A(0,A(0,A(0,1))) = A(0,A(0,2)) = A(0,3) = 4$  That's 6 steps, so if we define $RA(m,n)$ as the number of recursions to calculate $A(m,n)$, then $RA(1,2) = 6$.  I made a short Excel macro to calculate $RA(m,n)$ for any given $m$ and $n$. The problem I have is that $RA(m,n)$ grows even faster than $A(m,n)$ and I quickly run out of stackspace. The table below shows $RA(m,n)$ for very small values of $m$ and $n$. The numbers shown for $RA(3,10)$ and $RA(4,1)$ are the number of steps before crashing!    A formula for $RA(1,n)$ is trivial, and it is relatively easy to see that $RA(2,n) = 5(n+1) + 4n(n+1)/2$.  So my questions: can anyone see an explicit formula for $RA(3,n)$? More generally, does a formula exist for $RA(m,n)$, in the same way that $A(m,n)$ can be expressed using Knuth arrows?   For the low values shown of $m$ and $n$, $RA(m,n)$ looks about order of magnitude larger than $A(m,n)$. Can we prove that $RA(m,n) &gt; A(m,n)$ for all $m,n$. Intuitively, it should be.  For bonus points, can $RA(m,n)$ be expressed in terms of $A(m,n)$? Confused about complex numbers I am confused about something: \begin{eqnarray}     (e^{2 i \pi})^{0.5} = (e^{2 i \pi \cdot 0.5})= e^{i \pi}=-1 \end{eqnarray} but \begin{eqnarray}      e^{2 i \pi}=1~ and~ 1^{0.5}=1  \end{eqnarray} where is my mistake??  thanks Positivity of a determinant I'm stuck to prove the following exercise : Given real numbers $x_1,\ldots,x_n$ and $y_1,\ldots,y_n$, show that  $$ \det(e^{\large{x_iy_j}})_{i,j=1}^n&gt;0 $$ provided that $x_1&lt;\cdots&lt;x_n$ and $y_1&lt;\cdots&lt;y_n$.   Any idea ? Trichromelibrary was suddenly downloading from Google Play Store From nowhere, my friend got a notification that com.google.android.trichromelibrary was downloading together with a progress bar.  Why? What is $\int_{-\pi}^\pi \cos(nx)\cos(mx)\,dx$? I'm pretty sure that there's a theorem that says that the Fourier coefficients of a sum of $\cos(nx)$ and $\sin(nx)$ 's are the coefficients of the sum itself.   I tried to prove that in the specific case of $f(x) = \sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$. In calculating the Fourier coefficients I got to calculating  $\cos(nx)\cdot\cos(mx)$, but I'm now stuck. Any tips would be appreciated! Proof that $A_n$ the only subgroup of $ S_n$ index $2$. I have what seems to me a very simple proof that $A_{n}$ is the only subgroup of $S_{n}$ of index 2.  Since I've seen other people prove it with what feel like really complicated methods (), I'm wondering if I've overlooked something.  Proof: Let $H$ be any subgroup of index 2 in $S_{n}$.  Then $H\cap A_{n}$ is a normal subgroup of $A_{n}$ and since $A_{n}$ is simple, $H$ is either the trivial or the improper subgroup of $A_{n}$.  If it's the trivial subgroup it doesn't have index 2 in $S_{n}$, and otherwise the theorem is proved. Verbless sentence Prove that a finite union of closed sets is also closed Let $X$ be a metric space.  If $F_i \subset X$ is closed for $1 \leq i \leq n$, prove that $\bigcup_{i=1}^n F_i$ is also closed.  I'm looking for a direct proof of this theorem.  (I already know a proof which first shows that a finite intersection of open sets is also open, and then applies De Morgan's law and the theorem "the complement of an open set is closed.")  Note that the theorem is not necessarily true for an infinite collection of closed $\{F_\alpha\}$.  Here are the definitions I'm using:  Let $X$ be a metric space with distance function $d(p, q)$.  For any $p \in X$, the neighborhood $N_r(p)$ is the set $\{x \in X \,|\, d(p, x) &lt; r\}$.  Any $p \in X$ is a limit point of $E$ if $\forall r &gt; 0$, $N_r(p) \cap E \neq \{p\}$ and $\neq \emptyset$.  Any subset $E$ of $X$ is closed if it contains all of its limit points. What is the difference between OpenOffice and the newly created LibreOffice? What is the difference between OpenOffice and LibreOffice? I understand that LibreOffice is another fork that is going to be developed. I have used OpenOffice since v1 and am wondering if there is a definite advantage to uninstalling and trying LibreOffice. How can I tell what RAM will fit my computer? If I want to add more RAM to my computer, how can I buy something compatible? What kinds are there? How do I determine which kind my computer will accept and how much I can add? I have tried some of the solutions posted elsewhere in the support forum, one seemed like it worked but after a steam update i found myself back at square one  I am a new Linux user and am already at a learning curve :)  Thanks for any help  Reading package lists... Done Building dependency tree        Reading state information... Done Some packages could not be installed. This may mean that you have requested an impossible situation or if you are using the unstable distribution that some required packages have not yet been created or been moved out of Incoming. The following information may help to resolve the situation:  The following packages have unmet dependencies:  libgl1-mesa-glx:i386 : Depends: libglapi-mesa:i386 (= 10.1.3-0ubuntu0.3)  unity-control-center : Depends: libcheese-gtk23 (&gt;= 3.4.0) but it is not going to be installed                         Depends: libcheese7 (&gt;= 3.0.1) but it is not going to be installed E: Error, pkgProblemResolver::Resolve generated breaks, this may be caused by held packages. Press return to continue: