Problem:  I have collection of many images or books which can belong to multiple categories like     C, algorithms, interview, database, OS, windows,    for images    icons, website headers, awesome, minimal, abstract, print design, photography   For songs  new songs, old songs, romantic songs, comedy songs, awesome songs, slow songs, fast songs   To properly organize all the items I have to create folder for each category and place duplicate copies or shortcuts (making shortcuts is also a hassle) in each folder. I want a alternate solution to this method  Current file browsers seem to ignore these type of cases or such functionality is hidden to normal user. (I am doubtful about the folder technology implementation in file browsers there should be something like tagging).  Does there exist a software or script or good folder structure to decrease my effort ??
I'm finishing professional school and over the years have acquired a pile of notes and articles that I'd like to hang onto.  I'd like to add to them and create sort of an archive of article and files that may be useful down the road.  I'd also like to organize this collection of files not only by simple grouping but also with tags.  I feel like that will make searching through them years later much easier.  Suggestions on software that would be good for this?  Just a general file manager, something that uses tags that can be attached to each file?
Any time I am inserting new point feature in QGIS, the Feature Attribute dialog box is popping up. Is there a way to prevent it? (I am willing to insert many points at once and this pop up box slows down the process). To make a JavaScript class with a public method I'd do something like:  function Restaurant() {}  Restaurant.prototype.buy_food = function(){    // something here }  Restaurant.prototype.use_restroom = function(){    // something here }   That way users of my class can:  var restaurant = new Restaurant(); restaurant.buy_food(); restaurant.use_restroom();   How do I create a private method that can be called by the buy_food and use_restroom methods but not externally by users of the class?  In other words, I want my method implementation to be able to do:  Restaurant.prototype.use_restroom = function() {    this.private_stuff(); }   But this shouldn't work:  var r = new Restaurant(); r.private_stuff();   How do I define private_stuff as a private method so both of these hold true?  I've read  a few times but it doesn't seem like "private" methods can be called by public methods and "privileged" methods can be called externally. Easiest way to prove that $2^{\aleph_0} = c$ $\aleph_0$ is the cardinality of the set of natural numbers, $\aleph_0 = |N|$. $c$ is the cardinality of the continuum, i.e. the set of real numbers $c = |R|$.  I know that $|P(A)| = 2^{|A|}$. This means that the cardinality of the power set of a set is 2 raised to the power of the cardinality of that set.  This basically means that to prove  $2^{\aleph_0} = c$, I need to prove $c = |P(N)|$. I've seen the Cantor diagonal argument, although I have no idea on how to use it to prove this.   Could anyone suggest ideas of the simplest way to prove that $2^{\aleph_0} = c$? I'm having some troubles understanding the solution (going from step $2$ to step $3$) of this linear congruence equation. Here's the problem with the solution:   Solve the congruence $2x + 11 \equiv 7 \pmod 3$.   First, we reduce all the coefficients $\pmod 3$:  $$2x + 2 \equiv 1 \pmod 3. \tag{1}$$  Next, we add $1$ to both sides, using the fact that $2 + 1 \equiv 0 \pmod 3$:  $$2x \equiv 2 \pmod 3. \tag{2}$$  Finally, we multiply both sides by $2$, using the fact that $2 · 2 = 4 \equiv 1 \pmod 3$:  $$x \equiv 1 \pmod 3. \tag{3}$$  That is, any number in the set $\{\ldots, −5, −2, 1, 4, \ldots\}$ will solve the original congruence.    Like I said my problem is the step between $(2)$ and $(3)$. If I multiply both sides of $(2)$ by $2$ I get:  $$4x \equiv 4 \pmod 3.$$  Now, I understand that $4$ is congruent to $1 \pmod 3$ so the RHS "simplifies" to $1 \pmod 3$ but I don't understand why the LHS "simplifies" to $x$. Can someone help me with this? I was editing a curve for hair, and pressed W to bring the specials menu and Subdivide it, but instead everything but the curve I had active dissapeared, and my cursor turned into a circle like the one from the circle select mode.  I'm guessing they changed the specials menu and I'm inside some unknown to me function, but I'm not sure how to turn everything back to normal. I won a Samsung Galaxy S6 Edge, signed into it with my Google account, played around with it, wiped it, and sold it.  Now the person can't get past setup because it's saying he must login with an account previously synced to the phone.   Is there anything I can do from my side to remove this lock? I remember that a couple of years ago a friend showed me and some other people the following expression:  $$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$  As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:  What is the value of this limit and how do we prove it rigorously?  (Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)  I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.   (1) The sequence $a_n$ satisfies $a_n&lt;1$ for all $n\ge 2$.  Note that $a_n&lt;1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n&lt;n^n$. I prove this second statement by induction. Observe that $1^2=1&lt;4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n&lt;n^n$ for some integer $n\ge 2$. Then  $$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}&lt;(n-1)n^n+n^{n+1}&lt;(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$  (2) The sequence $a_n$ is increasing for all $n\ge 2$.  We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)  (2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By , we may find:  $$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$  A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:  $$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$  By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:  $$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}&lt;\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$  (In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.  I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.  Thanks! Fatal Error: Allowed Memory Size of 134217728 Bytes Exhausted (CodeIgniter + XML-RPC) I have a bunch of client point of sale (POS) systems that periodically send new sales data to one centralized database, which stores the data into one big database for report generation.  The client POS is based on PHPPOS, and I have implemented a module that uses the standard XML-RPC library to send sales data to the service. The server system is built on CodeIgniter, and uses the XML-RPC and XML-RPCS libraries for the webservice component. Whenever I send a lot of sales data (as little as 50 rows from the sales table, and individual rows from sales_items pertaining to each item within the sale) I get the following error:  Fatal error: Allowed memory size of 134217728 bytes exhausted (tried to allocate 54 bytes)   128M is the default value in php.ini, but I assume that is a huge number to break. In fact, I have even tried setting this value to 1024M, and all it does is take a longer time to error out.  As for steps I've taken, I've tried disabling all processing on the server-side, and have rigged it to return a canned response regardless of the input. However, I believe the problem lies in the actual sending of the data. I've even tried disabling the maximum script execution time for PHP, and it still errors out. Why does $x$ divided by zero not equal $x$? Why does $x$ divided by zero not equal $x$? After all, $x$ is not being divided by anything. TikZ arrow emanating from node at an angle for a distance How do I draw a line of length l and at an angle of a from a circular node N in TikZ? How do you access browser history? Some e-Marketing tools claim to choose which web page to display based on where you were before. That is, if you've been browsing truck sites and then go to Ford.com, your first page would be of the Ford Explorer.  I know you can get the immediate preceding page with HTTP_REFERRER, but how do you know where they were 6 sites ago? calculating resistance for LED strips to use as parking lamps and turn signals I'm installing LED strips in my car's tail lamps to convert it from conventional bulbs to LEDs. I have a understanding of what I'm doing but not the complete knowledge to make the circuit work correctly without doing some guessing so I figured I would ask here.   I'm planning on working with and I need a length of 16", 15" and 14" to make then look correct in my housings. When I did the math the 19.7" strips have about 1.5 LEDs per inch and need to be segmented into 3 LED groups so I may have to get creative with the placement.   Ffom what I understand I need 2 diodes and a resistor. One diode on the parking lamp power, and one on the turn signal/brake power and then a resistor on the parking lamp  wire to step down the voltage enough to make them dim enough to use as a parking lamp and distinguish it from the brighter turn/brake lamps.  Does anyone know what size resistor and what size diode to use in this project? The voltage, with the car running, is around 13.8V to the rear lamps. I figure it should be running around 7V to the LEDs to light them as parking lamps.  Also will leaving out a cluster of 3 LEDs change the strips intensity to where I'll need different resistors for each strip? Is the set of real matrices diagonalizable in $\mathcal M_n(C)$ dense in the set of all companion matrices? Let $\mathcal E=\{A \in \mathcal M_n(\mathbb R): A \text{ is of companion form} \}$. That is, $A$ has the following form \begin{align*} A=\begin{pmatrix}  0 &amp; \cdots &amp; 0&amp; -a_{0} \\  1 &amp; \cdots &amp; 0 &amp; -a_{1}\\ \vdots &amp;\ddots &amp; \vdots &amp;\vdots \\ 0 &amp;\cdots &amp; 1 &amp; -a_{n-1} \end{pmatrix}. \end{align*} Let $\mathcal D = \{A \in \mathcal M_n(\mathbb R): A \text{ is diagonalizable in $\mathcal M_n(\mathbb C)$}\}$. We know $\mathcal D$ is dense in $\mathcal M_n(\mathbb R)$. I am wondering whether $\mathcal E \cap \mathcal D$ is dense in $\mathcal E$. The set $\mathcal E$ is closed. It is not clear to me whether this statement is true. Why do RAW images in Darktable have a lot of noise? How does the wording "for each" work? Some cards in magic has something happen for each {something}. For instance:  makes you gain 2 life for each creature you control. Tapping  adds one green mana to your mana pool, and you gain 1 life for each non-land card revealed.  How is this life gained: all at once or "one bit" at a time? Say I tap Selvala and 1 land, and 2 non-lands are revealed: do I gain 2 life, or do I gain 1 life two times? Would  have one or two +1/+1 counters placed on it if I had it in play at that point? Possible Duplicate:          How to download a Youtube video from command line?  Mac OS X and/or Ubuntu Linux. Optionally, i'd like to save it in .avi or .mkv format. Can the program counter on Intel CPUs can be read directly (that is without 'tricks') in kernel mode or some other mode? How do I simplify: $$\sum_{r = 0}^{\lfloor \frac{k}{2} \rfloor} \dbinom{k}{2r} \cdot \dbinom{n-k}{k-2r}?$$ Basically, the sum is: $\dbinom{k}{0} \cdot \dbinom{n-k}{k} + \dbinom{k}{2} \cdot \dbinom{n-k}{k-2} + \dbinom{k}{4} \cdot \dbinom{n-k}{k-4} \cdots.$ This seems very similar to Vandermonde's Identity but because we skip two everytime instead of one, I don't see how I can apply it to this problem. What are some good examples of a function  $f : \mathbb{R} \to \mathbb{R}$ where its derivative is equal to its inverse? I attempted to find a monomial that satisfied it by starting with $f(x) = ax^b$ and showing that $f^{-1}(x) = f'(x) \implies b-1=\frac{1}{b} \implies b=\phi$ and got $$f(x) = \frac{x^\phi}{\sqrt[\phi]{\phi}}$$ Which seems to work according to WolframAlpha, but I'm having trouble double-checking it. Any other ideas? How to upload files to server using JSP/Servlet?