I'm new to android application development. I'm on course of developing my first app which displays the time table of the selected class/faculty on the selected day(Mon,Tue etc.,). I'm all done with UI. Now my problem is how &amp; where to write the code for the entire database insertion (as it should be loaded from the beginning). Can any one help me with this in a detailed manner as i'm new to this field ?
If your application requires a database and it comes with built in data, what is the best way to ship that application? Should I:   Precreate the SQLite database and include it in the .apk? Include the SQL commands with the application and have it create the database and insert the data on first use?   The drawbacks I see are:   Possible SQLite version mismatches might cause problems and I currently don't know where the database should go and how to access it. It may take a really long time to create and populate the database on the device.   Any suggestions? Pointers to the documentation regarding any issues would be greatly appreciated.
C# Mono - Low Level Keyboard Hook Inner product on $C(\mathbb R)$ With axiom of choice it is possible to construct an inner product on $C(\mathbb R)$. My question is, is it possible to explicitly construct an inner product on $C(\mathbb R)$? I.e. to give a closed formula to calculate the inner product? I know it is straight-forward to write down a scalar product using a Hamel basis. This is not the answer I am looking for. This question came to me, when a student asked me in the lecture today 'whether there are vector spaces without inner products'. So I tried to find scalar produces for function spaces. I think I managed to write one down for $L^1((0,1))$. But I failed to construct one for $C(\mathbb R)$. iOS 9 Screen Recording in El Capitan I used to record iOS devices through QuickTime, however now it's only recording audio (not sure is this linked to OS X 10.11 or iOS 9).  How can I fix this (without third party sofrware)? While I was reading the book Exploratory examples for real analysis, I came across this:     to show that : $$\lim_{x\rightarrow x_o}\frac{1}{h(x)} = \frac{1}{\lim_{x\rightarrow x_o}{h(x)}} = \frac{1}{M}$$      Discussion:Let $\epsilon \gt 0$.we want to find $\delta \gt 0$ such that       if  $0 \lt |x-x_0| \lt \delta $,then $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|\lt \epsilon$.      In order to get expression involving $|h(x)-M|$,we first find the common denominator:        $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|=\Big|\frac{M-h(x)}{M~h(x)}\Big|$      $~=\Big|\frac{h(x)-M}{|M|~|h(x)|}\Big|$.      As we have to deal with variable term $|h(x)|$ in denominator.We can construct bound .Let $\epsilon =\frac{|M|}{2}$ ,since lim$_{x\to x_o}h(x)=M$,there exists $\delta_1\gt 0~~$ s.t.       if $0 \lt |x-x_0| \lt \delta_1 $,then $|h(x)-{M}|\lt \frac{|M|}{2}$.      As you study steps below,identify that step where $h(x)\neq 0$ for all $0 \lt |x-x_0| \lt \delta_1 $ plays a critical role:      $|M|=|M-h(x)+h(x)|$      $~~~~~~\leq |M-h(x)|+|h(x)|$      $~~~~~~= |h(x)-M|+|h(x)|$      $~~~~~~\lt \frac{M}{2}+|h(x)| \implies$      $\frac{|M|}{2} \lt |h(x)| \implies$      $\frac{1}{h(x)}\lt \frac{2}{|M|}$,where $0 \lt |x-x_0| \lt \delta_1 $.      Now that we've bound for $1/h(x)$,we can apply the definition to term $|h(x)-M|$.      Since     lim$_{x\rightarrow x_o}h(x)=M$,we can find $\delta_2 \gt 0$ such that,       $~~~~~~~~~~~~~~~~~~~~~~~~~~~~$if  $0 \lt |x-x_0| \lt \delta_2$, then $|h(x)-M|\lt \frac{|M|^2 \epsilon }{2}$   The thing I can't understand is for what purpose did the author work with such complicated expression involving $\epsilon $ in the end of the discussion? Can I use old Minolta AF lenses on my new Canon? I just recently bought my first DSLR, a Canon Rebel t2i/550d. I use it mostly for video and it works wonderfully. I bought the kit lens with it, but I am looking to expand my lens collection. I found a set of old Minolta lenses from my dad, and I was wondering if I could use them on my new camera. The lenses are: a Sigma UC zoom 28-105 f4-5.6, a Minolta Maxxum zoom 70-210mm, a Minolta Maxxum zoom 28-58mm and a Kalimar 500mm lens. I know about the crop factor and that the auto focus won't work, but that does not bother me. Is there an adapter for Minolta AF/Sony to Canon EF mount? Any help is appreciated. Thanks! Configuring dpkg stops processing when calling an HTTP request to Adobe Flash Player Write access without read access Is it possible for a user to have a write access to a file and not be able to read it? How is it possible?   I tried the following commands:  debianbox@debian:~/posix/io$ touch filetest debianbox@debian:~/posix/io$ ls -l filetest -rw-r--r-- 1 debianbox debianbox 0 14 oct.  03:10 filetest  debianbox@debian:~/posix/io$ echo "Hello World" &gt; filetest debianbox@debian:~/posix/io$ cat filetest Hello World  debianbox@debian:~/posix/io$ chmod u-r filetest debianbox@debian:~/posix/io$ cat filetest cat: filetest: Permission forbidden  debianbox@debian:~/posix/io$   As you can see here, I have write but not read access on this file. How can this be possible? Is this considered as a bug? If not, in what situation would this be useful? Can't Use /Summon Command $X$ is Hausdorff if and only if the diagonal of $X\times X$ is closed Let $X$ be a topological space. The diagonal of $X \times X$ is the subset $$D = \{(x,x)\in X\times X\mid x \in X\}.$$   Show that $X$ is Hausdorff if and only if $D$ is closed in $X \times X$.   First, I tried to show that $X \times X \setminus D$ is open using the fact that $X \times X$ is Hausdorff (because $X$ is Hausdorff), but I couldn't find an open set that contains a point outside $D$ and is disjoint to it... Computing the Lie bracket on the Lie group $GL(n, \mathbb{R})$ Consider the Lie group $GL(n, \mathbb{R})$. Since $GL(n, \mathbb{R})$ is an open subset of the space $M_{n,n}(\mathbb{R})$ of $n \times n$ matrices, we can identify the tangent space (Lie algebra) $T_{e}GL(n, \mathbb{R})$ with $M_{n,n}(\mathbb{R})$, where $e$ is the identity matrix. I'm trying to understand the following:  For all $x, y \in M_{n,n}(\mathbb{R}) = T_{Id}GL(n, \mathbb{R})$ we have $[x, y] = xy -yx$ where $[\cdot, \cdot]$ is the Lie algebra bracket.  According to the question , this seems to be the case for matrix Lie groups in general.  So far, I've worked out the following: Let $x \in M_{n,n}(\mathbb{R}) = T_{Id}GL(n, \mathbb{R})$ and let $X \in \Gamma^{inv}(TG)$ be the left invariant vector field corresponding to $x$, i.e. with $X(e) = x$. By openness, we can choose $\epsilon &gt; 0$ such that the curve $(-\epsilon, \epsilon) \ni t \mapsto e + tx$ through the identity is contained in $GL(n, \mathbb{R})$. For $g \in GL(n, \mathbb{R})$ let $L_g$ be the left multiplication by $g$ in $GL(n, \mathbb{R})$. Using left invariance of $X$ calculate: $$X(g) = ((L_g)_{*}X)(g) = D_eL_g(X(e)) = D_eL_g(x) = \frac{d}{dt}|_{t=0}g(e+tx) = gx$$ Now, the above implies for $f \in C^{\infty}(GL(n, \mathbb{R}))$ and $g \in GL(n, \mathbb{R})$: $$X(g)(f) = (D_gf)(gx)$$ where we identify the tangent vector $X(g)$ with the corresponding derivation at $g$.  Now how do I calculate $[x,y] \stackrel{def.}{=} [X, Y](e) \stackrel{\text{WHY?}}{=}xy-yx$ from this? Difference in System. exit(0) , System.exit(-1), System.exit(1 ) in Java Better to buy an older high-level camera or newer entry-level camera? What are the pros and cons of buying an older higher-level camera if the price is about the same as a new entry-level one? Why does my regular expression work in X but not in Y? I wrote a regular expression which works well in a certain program (grep, sed, awk, perl, python, ruby, ksh, bash, zsh, find, emacs, vi, vim, gedit, …). But when I use it in a different program (or on a different unix variant), it stops matching. Why? Do I need a passport to visit Edinburgh in Scotland from elsewhere in the UK? I'm visiting Scotland for the first time this week, Edinburgh. I've lived in the UK for over 10 years but my nationality is Spanish. I only have a Spanish passport.  I'm also traveling by train.  Do I need a passport to enter Scotland and stay in my booked hotel? I am trying to draw a partial box around one or more words within a paragraph--that is, a rectangle with one or more sides missing. I have read the documentation for the fancybox package and it doesn't seem to be able to do what I want. I was able to produce thus ugly hack:    That's just a vertical bar | followed by a word with \overline and \uline. The trick is to connect the vertical and horizontal lines so that you have a rectangle missing just one side.  Is there an easy, general way to do this, perhaps with Tikz? I don't know enough about that package to be able to do this myself, but I'm wondering if there might be other solutions too. How can I edit two materials simultaneously? Whenever I open a new shading window, whatever I do in one is just mirrored on the other. If I click a node in one window, the same node and material get selected in the other. I can't edit two shaders at once. How can I edit two shaders simultaneously? In The Matrix Revolutions, when Neo and Trinity are trying to get to the Machine City, when the Machines launch weapons against them, Neo is able to destroy some of the incoming weapons. This is in the real world at this point, not in the Matrix.  How does Neo have any powers against the machines outside of the Matrix? Addendum: I'm looking for in universe answers only.  I would think that would be obvious, but I'm adding this after an answer that is essentially, &quot;It's a movie...&quot; Debian : Verify disk encryption password for already mounted system I'm trying to load data from different tables using left outer joins into a QgsVectorLayer. The approach I'm following so far is the one from the developers cookbook . The underlying database is postgres. The query returns correct data, including a geometry column, however I can't figure out how to get the join inside the query to work with . As the table the data come from needs to be specified, is it even possible to do so?  Would approaching the problem using psycopg2 be more suited, and if so, is there a way to create a layer from the result set? The only way I can think of is creating an empty layer and adding each feature individually by looping over the result set. But possibly there is an easier way?  Sadly, using the builtin DB Manager is not an option. The layer has to be added programmatically. Getting a standard form quadratic from a set of points ($3$ points) I came home from school today, pulled out my homework, now I'm stumped.  I don't want the answer, I just want to know how to do it.  Here is the question that I'm reading:  Determine a quadratic function in standard form for the set of points $(x_1,y_2)(x_2,y_2)(x_3,y_3)$  Where the $x$'s and $y$'s represent the points.  Here are the points I am working with (remember, I don't want the answer, I just want to know how to do it, there are lots of questions on this so I want to be able to know how to do it).     $(-1, 2), (0, 1), (1, -4)$   Thanks!