Laptop was slow so I tried to be smart and do factory reset. Picked the option to delete hard disk including partition without backing up. On laptop, laptop ask for system file. Don't have it.  Don't know what to do please help. I bet I deleted windows and also don't have the disk.
Recently, I accidentally deleted a partition off of my external hard disk. I had a copy of Minitool Power Data Recovery left over when the same drive had become RAW about a year ago. When I selected lost partition recovery, it immediately identified my deleted partition and allowed me to access and recover the files that were there.  My question is this. Is there any way to change the deleted partition back to NTFS without having to have a drive that you need to recover the files to and then move them back after formatting, as I don’t have any space left on either the computer or on my other external hard drive, so that I don’t need to buy a new one. The files on the disk were really important.  I am using Windows 7 Home Premium.  I appreciate any help in solving this crisis.
Bypass BIOS password set by faulty Toshiba firmware on Satellite A55-S1065 laptop? 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! With powershell get exactly the same application list as in Add/Remove programms There are various methods to find installed applications on the system but to my knowledge none returns exactly the same list as control panel item Add Remove Programs. Some of the ways I tried are:   WMI via Win32_Product and Win32Reg_AddRemovePrograms Registry, both x32 and x64 branches   For instance, I have 217 entries in AR and   gwmi  Win32_Product   returns 261 items.  gp HKLM:\Software\Microsoft\Windows\CurrentVersion\Uninstall\*, HKLM:\SOFTWARE\Wow6432Node\Microsoft\Windows\CurrentVersion\Uninstall\* | ? {![string]::IsNullOrWhiteSpace($_.DisplayName) } | select DisplayName   returns 119 items.   (gwmi Win32Reg_AddRemovePrograms) + (gwmi Win32Reg_AddRemovePrograms64) | sort DisplayName -Unique    returns 107 items.  I tried to nail it down with more complex filter that included InstallLocation, InstallSource, InstallDate etc without success. Gauss's mean value theorem problem Suppose that  $f(z)$ is analytic and $f(z)\neq0$  on the disc $|z|&lt;1$ .  Show that for $0&lt;r&lt;1$,  $$ \exp\left[\frac{1}{2\pi}\int_{0}^{2\pi}{\log}|f(re^{i\theta})|\,d\theta\right] = |f(0)| $$  Any other theorems needed to solve this problem? I don't have any idea just with "Gauss's MVT"  Thank you in advanced. Subgroups of Symmetric groups isomorphic to dihedral group Is $D_n$, the dihedral group of order $2n$, isomorphic to a subgroup of $S_n$ ( symmetric group of $n$ letters) for all $n&gt;2$? Specify CSS class names on SVG paths using Illustrator Is there a way to edit SVG files in Illustrator where you can specify a CSS class for each path?  I already know in Illustrator that if you give the layer name an actual name, Illustrator will use that name as the path's ID, which is fine if you don't plan on reusing the icon more than once on the same page.  My current workaround is to just use the IDs method, but then later convert the IDs to classes in my code editor, but it's pretty annoying to have to do every time I generate my SVG sprite.  If that's currently not possible in Illustrator, are there any other apps that will allow you to specify that? Or maybe a Grunt or Gulp package?  It looks like you might be able to do it with Inkscape with a , so I might look at that if there's no other good solutions out there. Is there a single word for something that is simultaneously not interesting and not boring? I am looking for a word that would describe a concept or activity, such as playing a game X, that is1 neither boring nor interesting. In other words, it should mean ‘neutral’ wrt intrestingness (~ boringness) but still comparable on the scale of intrestingness to both ‘boring’ and ‘interesting’.  An example sentence would be  As far as I'm concerned, long-distance running is [insert word]; it is not boring but it's also not interesting.  Criteria2   Has to be unambiguous3, and may consist of two words at most. Has to be a single word. Should be formal in the sense that it is included in Cambridge Dictionary, or Merriam Webster, or both. Should not contain negating prefixes, such as ‘un-’.   Does such a word exist in the English language?  For readers' convenience, please state which criteria you believe your word suggestion has met. You may do so, e.g., by adding a disclaimer at the top or bottom of your answer. Simply writing (1., 2.) is sufficient.        1 Assume a ternary system of intrestingness.   2 If not all criteria can be met, you may relax  the requirements from bottom to top, in the following order: 4. -> 3.  -> 2. -> 1. Answers with the smallest number of relaxed criteria in this order will be preferred in accepting. For example, fulfilling  (1., 2., 3.) is better than achieving (1., 3., 4.). Situations where a  higher priority ordering fulfills fewer criteria than a lower priority  vector, such as (1.) vs (3.,  4.), will be assessed on a case-by-case  basis.   3 Words such as ‘unappealing’ thus even fail  criterium 1 because in a three-valued system ‘unappealing’ ~ ‘not  appealing’ ~ ‘not interesting’ can be taken to mean either ‘boring’ or  ‘neutral wrt intrestingness’. Why do we use "./" (dot slash) to execute a file in Linux/UNIX? Why do we use ./filename to execute a file in linux?  Why not just enter it like other commands gcc, ls etc... How to access HTML element without ID? For instance in the snippet below - how do I access the h1 element knowing the ID of parent element (header-inner div)?  &lt;div id='header-inner'&gt;     &lt;div class='titlewrapper'&gt;        &lt;h1 class='title'&gt;        Some text I want to change       &lt;/h1&gt;     &lt;/div&gt;  &lt;/div&gt;   Thanks! Better explanation of the common general relativity illustration (stretched sheet of fabric) Is it possible to check who is entering your website in PHP. I have a web application ( written in PHP) that should only allow users entering from some particular websites. Is it possible to get the referral websites by examining the _Request object? If yes, how? How does transvoxel algorithm work and how can I implement It? Why can a 1 element set be a member of another set but not a subset of it? I have came across this in a textbook:     $\{2\}\nsubseteq\{\{2\}\}$ but $\{2\}\in\{\{2\}\}$   I understand that $\{2\}$ is an element (member) of the other set but considering $\{2\}$ is a set itself, why is it not a subset? I've recently set up the compiz cube again, and I was very happy to find that that flickering window glitch is gone! However, the cube reflection and deformation plugins are also gone.  I am running a fresh 12.10 install, and I have every compiz plugin package installed (including compiz-plugins-extra, which supposedly contains everything that compiz-fusion-plugins-extra package used to have), but those two plugins are still missing. I haven't been able to find anyone with the same problem.   Any ideas? (thanks!) My windows directory is huge Is 20 watts of electricity dangerous? I have a few circumstances which invlove someone being shocked with 20 watts of electricity and whether it would be deadly. So here they are:   Would 1000 volts at 20 miliamperes of AC (2MHz frequency) be dangerous or deadly? Would 1000 volts at 20 miliamperes of DC be dangerous or deadly? Would 10 Kilo volts at 2 miliamperes of AC (2MHz frequency) be dangerous or deadly? and Would 10 Kilo volts at 2 miliamperes of DC be dangerous or deadly?   Thanks for everyones help. Why is using a wild card with a Java import statement bad? Performance of try-catch in php What kind of performance implications are there to consider when using try-catch statements in php 5?   I've read some old and seemingly conflicting information on this subject on the web before. A lot of the framework I currently have to work with was created on php 4 and lacks many of the niceties of php 5. So, I don't have much experience myself in using try-catchs with php. Question from HNQ widget on Android doesn't load, showing last viewed question instead I have found myself in an uncomfortable situation after clicking the Hot Network Question widget.   I click the widget, it opens Stack Exchange app and doesn't refresh. That causes me seeing the question I haven been reading a while ago. Fact is that I haven't killed the app since using it last time.   Is there any known workaround to prevent this bug occurring every time I want to read a question opened from a widget? Killing the app/deleting data every time is not a convenient option.   It happens on all Android devices, so device specs aren't relevant, I guess.   Edit: it isn't reproducible 100% of the time. Sometimes it works. Using Vim I often want to replace a block of code with a block that I just yanked.  But when I delete the block of code that is to be replaced, that block itself goes into the register which erases the block I just yanked. So I've got in the habit of yanking, then inserting, then deleting what I didn't want, but with large blocks of code this gets messy trying to keep the inserted block and the block to delete separate.  So what is the slickest and quickest way to replace text in Vim?   is there a way to delete text without putting it into the register? is there a way to say e.g. "replace next word" or "replace up to next paragraph" or is the best way to somehow use the multi-register feature?