Read High voltage from PIC16F
How to read high voltages on microcontroller?
Let $g_n= {2^2}^n +1 $. Prove $g_0 · g_1 · · · g_{n−1} = g_{n} − 2$.    I'm not sure how to start this proof, if it should be done algebraically or if I should try to use a proof by induction. Electrical outlet on a dimmer switch? I live in an older house and have some outlets in our kitchen that don't make sense to me. We have one outlet that is tied to a dimmer switch on the opposite side of the room. I can detect that the outlet is hot using my voltage detector (when the dimmer switch is all the way on), but it pulses. I get about 6-7 beeps, then it skips one, beeps 6-7 more times, skips, continue. My GFCI tester shows an open hot. The outlet doesn't work.  I'm not very experienced with electrical work but this sounds highly unusual to me. Where should I start with fixing the issue? Why are bash tests so picky about whitespace? What can cause a question to be "bumped" to the first page in the Active tab of a Stack Exchange site? (Changing the "Last Active" date.)  . Create a tar archive split into blocks of a maximum size Do there exist an infinite number of integer-solutions $(x,y,z)$ of $x^x\cdot y^y=z^z$ where $1\lt x\le y$? Question : Do there exist an infinite number of integer-solutions $(x,y,z)$ of $x^x\cdot y^y=z^z$ where $1\lt x\le y$ ?  Motivation : After struggling to find a solution, I've just got one solution, which is  $$(x,y,z)=(1679616, 2985984, 4478976).$$  In the following, I'm going to write how I got this solution.  Letting $d$ be the greatest common divisor of $x,y,z$, we can represent $$x=ad, y=bd, z=cd$$ where $a,b,c$ are coprimes with each other. Then, we get $$d^{a+b-c}\cdot a^a\cdot b^b=c^c.$$  In the following, let's consider without the condition $x\le y$. Here, I suppose  $$a=2^m, b=3^n, a+b-c=1.$$  (As a result, this supposition works.)  Then, we get $$d=\frac{c^c}{2^{ma}\cdot 3^{nb}}.$$  Hence, letting $c=2^k\cdot 3^l$, if  $$kc\ge ma=m\cdot 2^m, lc\ge nb=n\cdot 3^n,$$ then $d$ is an integer.  Since $(m,n)=(4,2)$ satisfies the above conditions, then we get $d=2^8\cdot 3^6=186624.$ Hence we can get  $$x=9d=2^8\cdot 3^8=1679616, y=16d=2^{12}\cdot 3^6=2985984, z=24d=2^{11}\cdot 3^7=4478976.$$ Note that here I interchanged $x$ and $y$.  P.S : I was surprised to get this solution because I got this almost by chance. So, I don't know the other solutions. If you have any helpful information, please teach me. How to run batch file without admin privillege? We completed development on our dev environment and I copied the files over from the dev server to the production server through FTP.   Then through back up and migrade module, I made a back up of the SQL database. I created a new database through Plesk with the same name, username and password. I then imported the SQL database into the new server through phpmyadmin. I go to the URL and get the Site Maintenance message (which is correct, I placed the site in maintenance mode before copying the files over) but cannot navigate to any sub pages. Generally I can add /user or /admin to the end of the url and get the login screen. However all I get are 404 errors. How can I buy apps in the App Store without a credit card? A simple stupid question that I should already know the answer to:  Is this true: $\left|z^5\right| = \left|z\right|^5$ where z is a complex number with a non-zero imaginary component.  Thanks in advance. I have a valid B1/B2 US visa expiring about a year from now. The passport it's in expired a few years ago, and I recently got a new one.   The passport number does not match, but the national ID number, mentioned in both, does (it used to be the same as your national ID, but not anymore)  Can I travel to the US carrying both passports? Or is it asking for trouble? (I read a few years ago that it was possible, but that might have changed)  Should I just get a new visa to avoid potential complications? Why do Lambertian Reflectors Privilege the Observer? Ambiguity of "quite" The adverb "quite" has the following meanings :     1: wholly, completely ("not quite   finished")      2: to an extreme : positively "quite   sure" —often used as an intensifier   with a "quite a swell guy" "quite a   beauty"      3: to a considerable extent : rather   ("quite near")   My understanding is that #1 and #2 conflict with #3. For example, when you say "She's quite right", do you mean "She's completely right" or "To a considerable extent, she's right"? Notice that in the latter case you're not saying that she is completely right.  Another example is "I'm quite sure". Does it necessarily mean "I'm completely sure"?   Does the word really have conflicting meanings that can possibly cause ambiguity? Or am I missing something? Can an observed event in fact be of zero probability? Can an observed event in fact be of zero probability?  Of course, I know that there exist non-empty events of zero probability. Mu question is the reverse: given that we have observed an event (and we have no other information about it, just the fact that it has been observed), is it possible that the event has in fact zero probability? Or does an observation necessarily mean that the probability is strictly positive?    Example context for the question:  Suppose $x_i$ (countably many) are i.i.d on $[0,1]$, but we do not know the distribution they come from. It may be uniform on $[0,1]$, may be discrete, may be any legitimate distribution (discrete or continuous). Just imagine we have some sort of a machine that shows us one by one a list of randomly drawn numbers between $0$ and $1$. We are comparing the observed numbers one by one to some special number chosen beforehand, for example $\frac12$. Now, given that at some iteration we have observed that special number at least once, does that mean that $\frac12$ has some positive probability under that (unknown) distribution? And if the probability can be zero, can we nevertheless say that we will necessarily observe $\frac12$ again later, if we continue the experiment ad infimum?  Also, disregard the "real world limitations" such as an inability to produce truly uniformly distibuted numbers, or rounding errors or any such thing. Is the wave function of a particle re-created after a measurement stops? Evaluating the limit of a sequence given by recurrence relation $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Is my solution correct? I have a Transcend Storejet external USB hard disk. This is not an SSD, but a mechanical disk with rotating plates. I have formatted the whole disk, without partitioning, as NTFS. I have used mkntfs tool under Linux.  When plugged into a Linux machine, the system sees a drive with two partitions (/dev/sdc /dev/sdc1 /dev/sdc2). However, as I know it is a partitionless disk, I can mount the whole device (mount -t ntfs /dev/sdc /mnt) and it works without any problems.  When plugged into a MS Windows XP machine, the systems sees a disk with two unformatted partitions, and it does not assign a drive letter to either the whole disk, nor to any of the partitions.  Would anyone know how to get MS Windows to mount my disk as a NTFS superfloppy?  I have already tried removing old USB mounting points and residual devices using 'DriveCleanup'. It did not help.  By the way, I also have an external Kingston USB SSD, also formatted as NTFS superfloppy, without partitioning. However, this one is recognized and mounted normally by MS Windows. Proof that all abelian simple groups are cyclic groups of prime order What exactly do the colors in htop status bars mean? Grub rescue - error: unknown filesystem I have a multiboot system set up. The system has three drives. Multiboot is configured with Windows XP, Windows 7, and Ubuntu - all on the first drive. I had a lot of unpartitioned space left on the drive and was reserving it for adding other OSes and for storing files there in the future.  One day I went ahead and downloaded Partition Wizard and created a logical NTFS partition from within Windows 7, still some unpartitioned space left over. Everything worked fine, until I rebooted the computer a few days later.  Now I'm getting:    error: unknown filesystem.   grub rescue&gt;   First of all I was surprised not to find any kind of help command, by trying:    help, ?, man, --help, -h, bash, cmd, etc.  Now I'm stuck with non-bootable system. I have started researching the issue and finding that people usually recommend to boot to a Live CD and fix the issue from there. Is there a way to fix this issue from within grub rescue without the need for Live CD?  UPDATE  By following the steps from , I was able to boot to initramfs prompt. But not anywhere further than that.  So far from reading the manual on , I was able to see my drives and partitions using ls command. For the first hard drive I see the following: (hd0) (hd0,msdos6) (hd0,msdos5) (hd0,msdos2) (hd0,msdos1)  I now know that (hd0,msdos6) contains Linux on it, since ls (hd0,msdos6)/ lists directories. Others will give "error: unknown filesystem."  UPDATE 2  After the following commands I am now getting to the boot menu and can boot into Windows 7 and Ubuntu, but upon reboot I have to repeat these steps.    ls ls (hd0,msdos6)/ set root=(hd0,msdos6) ls / set prefix=(hd0,msdos6)/boot/grub insmod /boot/grub/linux.mod normal   UPDATE 3  Thanks Shashank Singh, with your instructions I have simplified my steps to the following. I have learned from you that I can replace msdos6 with just a 6 and that I can just do insmod normal instead of insmod /boot/grub/linux.mod. Now I just need to figure out how to save this settings from within grub itself, without booting into any OS.  set root=(hd0,6) set prefix=(hd0,6)/boot/grub insmod normal normal   UPDATE 4  Well, it seems like it is a requirement to boot into Linux. After booting into Ubuntu I have performed the following steps described in the :  sudo update-grub sudo grub-install /dev/sda   This did not resolve the issue. I still get the grub rescue prompt. What do I need to do to permanently fix it?  I have also learned that drive numbers as in hd0 need to be translated to drive letters as in /dev/sda for some commands. hd1 would be sdb, hd2 would be sdc, and so on. Partitions listed in grub as (hd0,msdos6) would be translated to /dev/sda6.  UPDATE 5  I could not figure out why the following did not fix grub:  sudo update-grub sudo grub-install /dev/sda   So I downloaded  based on an answer from  post. That seemed to do the trick after I picked the "Recommended Repair (repairs most frequent problems)" option.