I see two possible solutions:   Put the answering form immediately below the question (in the collapsed state by default) Put a copy of the question near the answering form.
Often when I edit one of my answers, it's because I want to better address the original question. The only problem is that I can't see the original question while I'm editing my answer!  I'd love to see the original question text beneath the "answer edit" form. This way I could scroll down slightly to read it if I wanted to, or just ignore it if not.  Current workaround is to click the 'return to answer' link and hope that you remember to launch it in a new tab/window, then tab back and forth.    Imported from this uservoice item:
Why will Ubuntu no longer measure file size unit as byte, megabyte, gigabyte, etc? I read on  that Ubuntu will no longer use the familiar file size units we all know by now (kB, MB, GB, TB) and switch to a different IEC standard (KiB, MiB, GiB, TiB). If this is true, I would like to know what's the reasoning behind this change, and the impact (if any) this change has, especially with multiplatform applications or applications run with Wine. Are actions and bonus actions interchangeable? The final state of 1000 light bulbs switched on/off by 1000 people passing by There are  1000  light bulbs and 1000 tutors. All light bulbs are off. Tutor 1 goes flipping light bulb 1,2,3,4... tutor 2 then flips 2,4,6,8... tutor 3 then 3,6,9...etc until all 1000 tutors have done this.  What is the status of light bulb 25? 93? 576?  Is there a general solution to find out the status of any light bulb?  How many light bulbs are on after all 1000 tutors have gone by?  Not homework but a UK university interview question. Please help. Showing that $\int_0^\infty x^{-x} \mathrm{d}x \leq 2$. This integral is very closely related to the sophmores dream that states  $$ \int_0^1 x^{-x}\mathrm{d}x = \sum_{n=1}^\infty n^{-n} = 1.27\ldots $$  For example here   Now I want to bound the integral, and showing that is less that 2.  For the interval $[0,1]$ a good bound is rewriting it to $\exp(x\log x)$ and using the expansion   $$ 1 - x \log(x) + \frac12 (-x \log(x))^2$$  but how does one handle $[1,\infty)$ ? In this answer here     gives bounds to the integral, but they are not tight enough.. So to taste my question again, how does one prove that   $$ \int_0^\infty \frac{\mathrm{d}x}{x^x} \leq 2 $$ Integral $\int\csc^3{x} \ dx$ I found these step which explain how to integrate $\csc^3{x} \ dx$. I understand everything, except the step I highlighted below.  How did we go from: $$\int\frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x}\,dx%$$ to  $$\int \frac{d(-\cot x + \csc x)}{-\cot x + \csc x} \quad?$$  Thank you for your time! $$ \int \csc^3 x\,dx = \int\csc^2x \csc x\,dx$$ To integrate by parts, let $dv = \csc^2x$ and $u=\csc x$. Then $v=-\cot x$ and $du = -\cot x \csc x \,dx$. Integrating by parts, we have: $$\begin{align*} \int\csc^2 x \csc x \,dx &amp;= -\cot x \csc x - \int(-\cot x)(-\cot x\csc x\,dx)\\ &amp;= -\cot x \csc x - \int \cot^2 x \csc x\,dx\\ &amp;= -\cot x\csc x - \int(\csc^2x - 1)\csc x\,dx &amp;\text{(since }\cot^2 x = \csc^2-1\text{)}\\ &amp;= -\cot x \csc x - \int(\csc^3 x - \csc x)\,dx\\ &amp;= -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx \end{align*}$$ From $$\int \csc^3 x\,dx = -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx$$ we obtain $$\begin{align*} \int\csc^3x\,dx + \int\csc^3 x\,dx &amp;= -\cot x \csc x + \int\csc x\,dx\\ 2\int\csc^3 x\,dx &amp;= -\cot x\csc x + \int\csc x\,dx\\ \int\csc^3x\,dx &amp;= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\csc x\,dx\\ &amp;=-\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\frac{\csc x(\csc x - \cot x)}{\csc x - \cot x}\,dx\\ &amp;= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{\csc^2 x - \csc x\cot x}{\csc x - \cot x}\,dx\\ &amp;= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{d(-\cot x+\csc x)}{-\cot x +\csc x}\\ &amp;= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\ln|\csc x - \cot x|+ C \end{align*}$$ This is a continued discussion from () by return count only.     Problem      Given four lists A, B, C, D of integer values, compute how many tuples   (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.      To make problem a bit easier, all A, B, C, D have same length of N   where \$0 \le N \le 500\$. All integers are in the range of \$-2^{28}\$ to \$2^{28} - 1\$   and the result is guaranteed to be at most \$2^{31} - 1\$.      Example:      Input:  A = [ 1, 2] B = [-2,-1] C = [-1, 2] D = [ 0, 2]       Output:  2       Explanation:      The two tuples are:         (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0   (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0      I'm wondering if there are any ideas to have a solution less than \$O(n^2)\$ time complexity.  Source code in Python 2.7,  from collections import defaultdict def four_sum(A, B, C, D):     sum_map = defaultdict(int)     result = 0     for i in A:         for j in B:             sum_map[i+j] += 1     for i in C:         for j in D:             if -(i+j) in sum_map:                 result += sum_map[-(i+j)]      return result  if __name__ == "__main__":     A = [1, 2]     B = [-2, -1]     C = [-1, 2]     D = [0, 2]     print four_sum(A,B,C,D) How are Dementors created/born? This is inspired by some discussions in .  In the Harry Potter universe, how are Dementors born?  Or are they artificially created by some magical means? Prove that $\nabla_{\mathrm X} \mbox{tr} (\mathrm A \mathrm X^{-1} \mathrm B) = - \mathrm X^{-\top} \mathrm A^\top \mathrm B^\top \mathrm X^{-\top}$ Prove that $$\nabla_{\mathrm X} \mbox{tr} (\mathrm A \mathrm X^{-1} \mathrm B) = - \mathrm X^{-\top} \mathrm A^\top \mathrm B^\top \mathrm X^{-\top}$$   My proof is below. I am interested in other proofs.    My proof  Let   $$f (\mathrm X) := \mbox{tr} (\mathrm A \mathrm X^{-1} \mathrm B)$$  Hence,  $$\begin{array}{rl} f (\mathrm X + h \mathrm V) &amp;= \mbox{tr} (\mathrm A (\mathrm X + h \mathrm V)^{-1} \mathrm B)\\ &amp;= \mbox{tr} (\mathrm A (\mathrm X ( \mathrm I + h \mathrm X^{-1} \mathrm V))^{-1} \mathrm B)\\ &amp;= \mbox{tr} (\mathrm A ( \mathrm I + h \mathrm X^{-1} \mathrm V)^{-1} \mathrm X^{-1} \mathrm B)\\ &amp;= \mbox{tr} (\mathrm A ( \mathrm I - h \mathrm X^{-1} \mathrm V + O (h^2)) \mathrm X^{-1} \mathrm B)\\ &amp;= \mbox{tr} (\mathrm A \mathrm X^{-1} \mathrm B) - h \, \mbox{tr} ( \mathrm A \mathrm X^{-1} \mathrm V \mathrm X^{-1} \mathrm B) + O (h^2)\\ &amp;= f (\mathrm X) - h \, \mbox{tr} ( \mathrm X^{-1} \mathrm B \mathrm A \mathrm X^{-1} \mathrm V ) + O (h^2)\end{array}$$  Thus, the directional derivative of $f$ in the direction of $\mathrm V$ at $\mathrm X$ is  $$D_{\mathrm V} f (\mathrm X) = - \mbox{tr} ( \mathrm X^{-1} \mathrm B \mathrm A \mathrm X^{-1} \mathrm V ) = - \mbox{tr} ( (\mathrm X^{-\top} \mathrm A^\top \mathrm B^\top \mathrm X^{-\top})^\top \mathrm V ) = - \langle \mathrm X^{-\top} \mathrm A^\top \mathrm B^\top \mathrm X^{-\top}, \mathrm V \rangle$$  and, lastly,  $$\nabla_{\mathrm X} \mbox{tr} (\mathrm A \mathrm X^{-1} \mathrm B) = - \mathrm X^{-\top} \mathrm A^\top \mathrm B^\top \mathrm X^{-\top}$$ Blender Nodes in Editor doesn't show preview Image I have a small problem with my blender lately. I want to make some compositing with my compositing nodes but the problem is, blender won't show me the preview Image. I have a Screenshot for you guys.    The preview Image itself is enabled, but it just won't show.  Thanks for your help in advance!  -Marten Using the simple example below, what is the best way to return results from multiple tables using Linq to SQL?  Say I have two tables:  Dogs:   Name, Age, BreedId Breeds: BreedId, BreedName   I want to return all dogs with their BreedName.  I should get all dogs using something like this with no problems:  public IQueryable&lt;Dog&gt; GetDogs() {     var db = new DogDataContext(ConnectString);     var result = from d in db.Dogs                  join b in db.Breeds on d.BreedId equals b.BreedId                  select d;     return result; }   But if I want dogs with breeds and try this I have problems:  public IQueryable&lt;Dog&gt; GetDogsWithBreedNames() {     var db = new DogDataContext(ConnectString);     var result = from d in db.Dogs                  join b in db.Breeds on d.BreedId equals b.BreedId                  select new                         {                             Name = d.Name,                             BreedName = b.BreedName                         };     return result; }   Now I realize that the compiler won't let me return a set of anonymous types since it's expecting Dogs, but is there a way to return this without having to create a custom type?  Or do I have to create my own class for DogsWithBreedNames and specify that type in the select?  Or is there another easier way? How to divide two 64-bit numbers in Linux Kernel? Some code that rounds up the division to demonstrate (C-syntax):  #define SINT64 long long int #define SINT32 long int  SINT64 divRound(SINT64 dividend, SINT64 divisor) {   SINT32 quotient1 = dividend / divisor;    SINT32 modResult = dividend % divisor;   SINT32 multResult = modResult * 2;   SINT32 quotient2 = multResult / divisor;    SINT64 result = quotient1 + quotient2;    return ( result ); }   Now, if this were User-space we probably wouldn't even notice that our compiler is generating code for those operators (e.g. divdi3() for division). Chances are we link with libgcc without even knowing it. The problem is that Kernel-space is different (e.g. no libgcc). What to do?  Crawl Google for a while, notice that pretty much everyone addresses the unsigned variant:  #define UINT64 long long int #define UINT32 long int  UINT64 divRound(UINT64 dividend, UINT64 divisor) {   UINT32 quotient1 = dividend / divisor;    UINT32 modResult = dividend % divisor;   UINT32 multResult = modResult * 2;   UINT32 quotient2 = multResult / divisor;    UINT64 result = quotient1 + quotient2;    return ( result ); }   I know how to fix this one: Override udivdi3() and umoddi3() with do_div() from asm/div64.h. Done right? Wrong. Signed is not the same as unsigned, sdivdi3() does not simply call udivdi3(), they are separate functions for a reason.  Have you solved this problem? Do you know of a library that will help me do this? I'm really stuck so whatever you might see here that I just don't right now would be really helpful.  Thanks, Chad Execute shell commands in Python Is any aspect of the supernatural testable? What level of proof is possible for the supernatural? What's the derivative of: $ \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$? let $ y=\displaystyle \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$, i'm really interesting to know how do I find :  $\displaystyle \frac{dy}{dx}$ ?.  Note: I have used the definition of derivative of the square root function but i don't succed .  Thank you for any help Should we have another site for frameworks (and alike) recommendations? There is . Is the site appropriate for stuff related to requesting frameworks and libraries? Or should we have another site for that (e.g. name it Frameworks Recommendations or stuff alike)?  I can ask stuff there like: I need a software satisfying W, X, Y, and Z, but I'd like -some day- ask like: I need a Python library satisfying W, X, Y, and Z. The audience may be (and perhaps will be) different.    Is that website on-topic for stuff like that? Or should we open another proposal for that?  No: It's not about asking for a giga-dump for subjectivish questions. The linked-to-dupe question was written and answered with an old mindset and without Software Recommendations in mind. Also it does not consider this specific kind of questions but all the crap in one place. So, no,  is not an appropriate dupe mark for this question. Changing default width of blocks in beamer How check with SOQL if a picklist value is available or not for a specified record type How check if a picklist value is available or not for a specified record type ( through SOQL)? Thanks in advance Find conditional expectation from a 3-dimensional random vector? I am looking for a method to process a remote sensing image and extract the crown areas of the individual trees from the image.   I have both visual wavelength areal imagery, and lidar data from the area. The location in question is a desert area, so the tree cover isn't as dense as a forest area. The resolution of the aerial imagery is 0.5 feet by 0.5 feet. The lidar resolution is approximately 1 x 1 feet. Both the visual data and the lidar come from a Pima County, Arizona dataset. A sample of the type of aerial imagery I have is at the end of this post.   This question  seems to be the same issue, but there does not seem to be a good answer there.  I can obtain a reasonable classification of the vegetation types (and information about the overall percent cover) in the area by using the Iso Cluster classification in Arcmap, but this provides little information on individual trees. The closest I have to what I want is the results of passing the output of the isocluster classification through the Raster to Polygon feature in Arcmap. The problem is that this method merges near by trees into a single polygon.   Edit: I probably should have included some more detail about what I have. The raw datasets I have are:    Full las data, and a tiff raster generated from it. Visual imagery (like the sample image shown, but covering a much wider area)  Manual direct measurements of a subset of the trees in the area.    From these I have generated:    The ground/vegetation classifications. The DEM/DSM rasters. A site (or scraper) is copying content from Stack Exchange. What should I do? of Stack Overflow, all content posted on Stack Exchange sites by their users (i.e. you wonderful people) has been provided to the whole universe under . For my fellow non-lawyers, that license basically means:   Anyone can use any Stack Exchange posts at any time without having to ask for permission Making money off of the copied content is permitted You don't even have to copy stuff from here verbatim; you can just use it as a starting point and make whatever edits you want There are just two rules you have to follow:   You have to provide . Simple links to the original post and author info are just fine. You have to  and allow other people to use your content, as long as they follow these very same rules. How meta!    (If you ever forget any of that, and want to refresh your memory, the license info is linked to in the footer of every page.)  There are, in fact, a lot of people who republish varying amounts of our content for assorted reasons. Unfortunately, there are some Stack Content Republishers Attributing Poorly and/or Excelling at Ranking (SCRAPERs, for short).  In this context, "attributing poorly" means any use that doesn't follow our attribution rules or make any other reasonable attempt at give credit. This can get pretty egregious; I've seen SCRAPERs that not only don't link back to SE originals, but also use fake author info and post dates to make it harder to find originals. By "excelling at ranking," I'm referring to copycat sites that end up higher in Google results than the original SE sites do for the same content. That's not necessarily wrong, but in some cases, it indicates inappropriate SEO hackery.  So, the question is: what can you do if you spot a SCRAPER?