I'm having trouble installing Oracle java jdk
How can I install Sun/Oracle's proprietary Java JDK 6/7/8 or JRE?
I would like a way to mark a question in the Unanswered "my tags" page as something I'm not interested in or able to answer The /unanswered/tagged?tab=mytags page gives a list of questions from my favorite tags that are currently considered "unanswered" - they have no up voted or accepted answers. I think it's good to put them in front of me easily, but even though they have one or more tags that I'm interested in, I may not have the knowledge, ability, or interest to write an answer to them. However, they will remain on this page until someone writes an answer.  I would like to be able to dismiss questions from views, like this one, that show me interesting questions. I may choose to share them on my social media to get them views and answers, but I don't want to continually see them. Instead, I want to be able to use views like unanswered/tagged?tab=mytags as a queue of things that could use my attention. I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ which are continuous and satisfy $f(x+y)=f(x)+f(y)$ I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $f(x+y)=f(x)+f(y)$. I know that there are other questions that are asking the same thing, but I'm trying to figure this out by myself as best as possible. Here is how I started out:  Try out some cases:  $x=0:$ $$f(0+y)=f(0)+f(y) \iff f(y)=f(0)+f(y) \iff 0=f(0) $$ The same result is for when $y=0$  $x=-y:$ $$f(-y+y)=f(-y)+f(y) \iff f(0)=f(-y)+f(y) \iff 0=f(-y)+f(y)\iff \quad f(-y)=-f(y)$$ I want to extend the result of setting $x=-y$ to numbers other that $-1$, perhaps all real numbers or all rational numbers. I got a little help from reading other solutions on the next part:  Let $q=1+1+1+...+1$. Then  $$f(qx)=f((1+1+...+1)x)=f(x+x+...+x)=f(x)+f(x)+...+f(x)=qf(x)$$ I understood this part, but I don't understand why this helps me find all the functions that satisfy the requirement that $f(x+y)=f(x)+f(y)$, but here is how I went on:  Thus  $$f(qx)=qf(x)$$ and it should follow that $$f \bigg (\frac {1}{q} x\bigg)= \frac{1}{q}f(x)$$ where $q\not =0$, then it further follows that  $$f \bigg (\frac {p}{q} x\bigg)= \frac{p}{q}f(x)$$ where $\frac{p}{q}$ is rational, and lastly it further follows that  $$f (ax)= af(x)$$ where $a$ is real. Thus functions of the form $f(ax)$ where $a$ is real satisfies the requirement of  $f(x+y)=f(x)+f(y)$.  I don't know how much of what I did is correct\incorrect, and any help would be greatly appreciated. Also is there any way that I can say that functions of the form $f(ax)$ where $a$ is real are the only functions that satisfy the requirement of  $f(x+y)=f(x)+f(y)$? Or do other solutions exist?  Again, thanks a lot for any help! (Hints would be appreciated, I'll really try to understand the hints!) If $G$ is cyclic, then $G/H$ is cyclic?   The proof I got goes like this: $G$ is cyclic, so $G=&lt;g&gt;$ for some $g\in G$. So any coset in $G/H$ would be of the form $Hg'=Hg^n$ for some $n$. So $Hg$ is an generator of $G/H$. Thus, $G/H$ is cyclic.  I might just be confusing myself, but we have only shown that $Hg'$ is in form of $(Hg)^n$. But what if we are missing some $n\in\mathbb{N}$? That is, there is no quotient group of the form $Hg^2$, for example.  To make myself a little bit clearer, I think what the above proof has done was showing that $\forall Hg'\in G/H, Hg'\in &lt;Hg&gt;$, thus $G/H \subset &lt;Hg&gt;$. I feel that this is not a complete proof. Strange black shading. Cannot fully light mesh. (Cycles) Here is the viewport in material view.  The node setup is just a PNG texture into a diffuse shader.    And this is in rendered view.  I've tried using environmental lighting as well as a directional light, and while the lit parts become brighter/washed out, the darker parts stay dark.  Normals are correct and there is no z-fighting.  Ambient Occlusion is not enabled.  Changing to Flat shading makes it worse.    Update:  Any modifier that alters the mesh (bevel, subdivide, etc) make the problem go away instantly. Find all natural numbers $n &gt; 1$ and $m &gt; 1$ such that $1!3!5!\cdots(2n - 1)! = m!$   I have been thinking about coming up with some inequalities which would narrow the possible range of pairs $(n, m)$, however the best I have been able to find so far are $m \ge 2n - 1$ and $m \lt n^2$, which is clearly not enough.  Update. The question has tag combinatorics as it is from a book about combinatorics, so there must be at least partly combinatorial solution. It has been said that RSA uses a modulus product of two prime numbers for security reasons. But does RSA even work correctly if we allow composite integers instead?  I think that the answer is "NO".  Define “work correctly” as:   allows encryption and decryption to be carried out as in normal RSA, yielding the original message (message padding and unpadding can remain unspecified, or absent as in texbook RSA); a security level can be given in terms of the key generation process (which may have alternate requirements to testing that $p$ and $q$ are prime). In most fiction I've read or encountered, the change occurs as a simple consequence of the lunar cycle. Though from what I know of Harry Potter, the change that takes place is due to sighting the full moon (as occurs in the third movie, Prisoner of Azkaban) where the gang, plus Lupin, Severus, Sirius and Pettigrew emerge from the Whomping Willow and upon sighting the full moon he begins to change.  Is this the trigger for werewolves in the Potterverse? If not, why did he (Lupin as an example, though for any werewolf in general) not change whilst in the shrieking shack that night?  Edit: This  addresses Lupin's change but I'm after more general information about werewolves. BibTeX handling of the Dutch "van" name prefix with natbib I want to produce the following with LaTeX/BibTeX:  In the running text:     ... we refer the reader to the work of Van Noort (2010).   In the bibliography:     Thomas van Noort. An important paper. 2010.   I have tried many tricks, but I don't seem to able to fulfill all the requirements:   "Thomas van Noort" has to appear sorted in the bibliography by the name "Noort". In the running text, the "Van" needs to have an uppercase 'V' In the bibliography, the "van" needs to have a lowercase 'v' I am using hyperref, and I want the whole "Van Noort (2010)" text to link to the reference.   Using plain \citet will generate:     ... we refer the reader to the work of van Noort (2010).   Note the lowercase 'v'; this is not acceptable, as per standard treatment of Dutch names. Setting the BibTeX name to Van Noort, Thomas makes the running text citation look ok, but then in the bibliography I get     Thomas Van Noort. An important paper. 2010.   Note the uppercase 'V'; this is also not acceptable, as it should only be uppercase if it is not preceded by the first name.  (1) is easy: I set the BibTeX author field to {\noopsort{Noort}}{van Noort}, Thomas. (2) and (3) I can fake by saying something like we refer the reader to the work of Van Noort \citeyear{...}, but this is not nice, and does not fulfill (4).  Anyone has any suggestions? I was under the impression that LaTeX and BibTeX were designed with these kind of things in mind, so maybe I'm missing something... (How) Can I use existing stapled electrical wire to fish new wire? I nicked an electrical cable while doing a plumbing repair and so need to fix it. I previously  about repairing the damage in place, but there are receptacles 6-10 feet away on either side of the damaged wire, so it may not be too difficult to just pull new wire. Most of the wire is run horizontally through the studs, which is great, but it will at least be stapled at both boxes (and possibly more). The horizontal run is ~18" up from the floor, while the receptacles are ~6". There is also an inside corner the wire is run through. Wire is 12/2.  Is it possible to use the existing wire to fish the new wire, even though the existing wire is stapled? Would I need to open up the wall to remove the staples?  I found a , but that pertains to telephone wires which sometimes use different staples than electrical wires. Book featuring a child learning from a crowdsourced AI book How can I check the version of the available package in the Ubuntu repositories without installing it? Are $_SESSION variables stored on the client or the server? Don't you think people need to know what they did wrong exactly so they can fix it and get unbanned from asking questions or posting answers? I have the following code: info = new System.Diagnostics.ProcessStartInfo(&quot;TheProgram.exe&quot;, String.Join(&quot; &quot;, args)); info.CreateNoWindow = true; info.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden; info.RedirectStandardOutput = true; info.UseShellExecute = false; System.Diagnostics.Process p = System.Diagnostics.Process.Start(info); p.WaitForExit(); Console.WriteLine(p.StandardOutput.ReadToEnd()); //need the StandardOutput contents  I know that the output from the process I am starting is around 7MB long. Running it in the Windows console works fine. Unfortunately programmatically this hangs indefinitely at WaitForExit. Note also this code does NOT hang for smaller outputs (like 3KB). Is it possible that the internal StandardOutput in ProcessStartInfo can't buffer 7MB? If so, what should I do instead? If not, what am I doing wrong? For-each over an array in JavaScript Lost credit card replaced with new card and new numbers. Credit score affected? I recently left my credit card at a bar so I reported it as lost to my bank, and my bank said I will get a replacement within a few days. For what it's worth, I picked my card up from the bar a few days later and made sure to cut it up.  The new credit card has arrived and has new numbers. Will this negatively affect my credit score?  VISA, Chase Bank, USA. My desktop look is empty so all I have is a nice wallpaper. My Windows taskbar has the "Taskbar Appearance" setting to "Auto-hide the taskbar". Although hidden, it leaves a thin top layer so you can still see it a bit, and where the pinned shortcuts are. Is it possible to totally have it hidden without a trace? This is a serious question. It really bugs m. Could normal users have an easier way to flag for migration? Please allow using nested expressions when filtering with tags Please support the use of nested expressions like  and (or ) when searching.  Looking here, on meta, at questions tagged   and at questions found with +search +tag +boolean I didn't see any information on this topic. What should I do if I have two libraries that provide functions with equivalent names?