Was running Windows 7 (64-bit). Downloaded Ubuntu, and booted it from USB Started installing Ubuntu OS. I chose the last option, do not replace Windows, do something else.  However, when I had to choose a partition, I ticked and formatted the one in which my Windows 7 was on.  After formatting, I cancelled installation.  So, I now have neither have windows nor Ubuntu.  When I switch it on, I just get the System Recovery options.   I cannot do that as I still have important files: pictures and videos (I am a photographer.)  Please tell me there is a way to get them back!
I have a Toshiba satellite A-200 laptop with a Vista OS on it with 4 NTFS partitions  (C:) Vista  (D:) Entertainment  (E:) Work  (F:) Sources   and I wanted to start using Ubuntu instead. So I tried it first from the live CD and everything was OK and all the partitions were shown and working and so I decided to install Ubuntu to replace Vista on the (C:) drive. After I did that I can no longer find my folders and files on the (D:), (E:), (F:) partitions and the only file system that is shown is one 198 GB although my HDD is 320 GB.  I can't access the lost data on the remaining 120 GB which I hope is still there and not totally lost  I am now working from the live CD but I am unable to install testdisk. Can I recover the Vista partitions by the product recovery CD to get my laptop back to the factory settings? Can I recover the NTFS partitions using a recovery program for Windows or will that make the problem worse?   I need these data badly as I don't have a backup for them.
Inverse function of $x^x$ How can I find the inverse function of $f(x) = x^x$? I cannot seem to find the inverse of this function, or any function in which there is both an $x$ in the exponent as well as the base. I have tried using logs, differentiating, etc, etc, but to no avail. Traces of all positive powers of a matrix are zero implies it is nilpotent Let $A$ be an $n\times n$ complex nilpotent matrix. Then we know that because all eigenvalues of $A$ must be $0$, it follows that $\text{tr}(A^n)=0$ for all positive integers $n$.  What I would like to show is the converse, that is,      if $\text{tr}(A^n)=0$ for all positive integers $n$, then $A$ is nilpotent.   I tried to show that $0$ must be an eigenvalue of $A$, then try to show that all other eigenvalues must be equal to 0. However, I am stuck at the point where I need to show that $\det(A)=0$.  May I know of the approach to show that $A$ is nilpotent? Show that, for all $n > 1: \frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n.$ I'm learning calculus, specifically derivatives and applications of MVT, and need help with the following exercice:      Show that, for all $n &gt; 1$ $$\frac{1}{n + 1} &lt; \log(1 + \frac1n) &lt; \frac1n.$$    I tried to follow the below steps in order to prove the RHS inequality:   Proving that $f &lt; g$ on $I$ from $a$ to $b$:  Step $1$. Prove that $f' &lt; g'$ on $\operatorname{Int}(I)$.  Step $2$. Show that $f(a) \leq g(a)$ or that $f(a^+) \leq g(a^+)$  Following the above steps, let $f(x) = \log(1 + \frac1x)$ and $g(x) = \frac1x$, for all $x &gt; 1$. One has  $$f'(x) =  -\frac{1}{x^2 + x} \quad \text{and} \quad g'(x) = -\frac{1}{x^2}.$$  We note that, for every $x &gt; 1$, $f'(x) &gt; g'(x)$. Moreover, $f(1^+) = \log(2) &lt; 1 = g(1^+).$   My problem is that I got the wrong inequality sign in Step $1$.    Looking at the solution in my textbook, the author suggests using the MVT but I don't know how to apply it in this case. Is it possible to fail an ability check on purpose? Here are the general guidelines regarding ability checks:     The GM calls for an ability check when a character or monster attempts an action (other than an attack) that has a chance of failure. When the outcome is uncertain, the dice determine the results.   In some cases, you could want to fail an ability check. If you are climbing a wall an want to drop off, you fail simply because you stop attempting to climb, but in other cases, success and failure are both tied to the actual action and may both have desirable outcomes.  Example 1: Spell scrolls  An evil adversary wants you to cast a spell scroll for him, but you have him figured out and want to destroy his spell scroll instead, seemingly by accident. The rules on spell scrolls read:     If the spell is on your class’s spell list but of a higher level than you can normally cast, you must make an ability check using your spellcasting ability to determine whether you cast it successfully. The DC equals 10 + the spell’s level. On a failed check, the spell disappears from the scroll with no other effect.   A failure means that you successfully destroy the spell scroll.  Example 2: Amulet of the Planes  You are in possesion of the Amulet of the Planes. Your party is being overrun by terrible monsters, and the only way to save your party is to sacrifice yourself by sending yourself and the monsters to another plane of existence.     While wearing this Amulet, you can use an action to name a location that you are familiar with on another plane of existence. Then make a DC 15 Intelligence check. On a successful check, you cast the Plane Shift spell. On a failure, you and each creature and object within 15 feet of you travel to a random destination. Roll a d100. On a 1-60, you travel to a random location on the plane you named. On a 61-100, you travel to a randomly determined plane of existence.   By walking up to the monsters and failing the check, you and the monsters would be sent to another plane of existence.  The question  The rules seem to assume that you always want to succeed when you perform an ability check. Sometimes, however, you could want to fail on purpose. A sensible DM would rule that if you are skilled enough to succeed the check, then you would undoubtly be able to fail it on purpose as well. What does the rules as written say -- if anything -- about this circumstance?  According to RAW, is it possible to attempt an ability check and intentionally fail? How does the current limiting resistor for an LED affect current and voltage drops? I'm having a bit of trouble understanding currrent limiting resistors in simple LED circuits. I know that I can determine the optimal resistor like so:  \$\displaystyle R=\frac{V_{s}-V_{f}}{I_{f}}\$  But I'm having a hard time understanding how this one value modifies the voltage and the current to the correct values for the LED. For example, if my calcuations for a super bright blue LED (with \$V_{f}\$ being 3.0-3.4&nbsp;V and \$I_{f}\$ being 80&nbsp;mA, and a voltage source of 5&nbsp;V) gives me 25&nbsp;ohms (using the lower bound of the forward voltage), that's fine. So the current throughout should be 80&nbsp;mA and the voltage drop for the resistor and LED should be 2 and 3 volts, respectively.  But what if I used a 100&nbsp;ohm resistor instead? Or any other value—how would I calculate the voltage drops and current? Would I assume one of them stays the same? What is the proof that the total number of subsets of a set is $2^n$? What is the proof that given a set of $n$ elements there are $2^n$ possible subsets (including the empty-set and the original set). If I'm floating in space and I turn on a flashlight, will I accelerate? Why I can't read a read only file? Explain proof by contradiction. How to show that $L^p$ spaces are nested? Suppose $1&lt;p_1&lt;p_2&lt;\infty$, then show $L^{p_1}[a,b] \supset L^{p_2}[a,b]$.  I was able to show $\|f\|_{p_1} \le \|f\|_{p_2} (b-a)^{1/p_1 - 1/p_2}$ but I'm not sure how to proceed from here. Mob Arena - Mob vs. Mob I want to make an arena where mobs will fight each other and a player can bet on one side or the other. For example: There are 2 skeletons in dyed armor; red and blue. Each color representing a team. When they spawn I want them to fight each and then when one dies I want it to announce which team is the winner and give the player the "money" due. I'm just not sure how to get the skeletons to attack each other without a player instigating it. Helping someone climb I know this is kinda stupid but what is the word for "helping someone climb by allowing him to stand on your hands" (to allow him/her see above some wall or climb something)? How can I add certain mobs to a team and make them passive to me if they would normally attack me? I know how to do it in 1.8, but how can I do it now in 1.16.4? Based on this question: , LVS may be the right solution for my problem.   But when I went to its homepage , I found that LVS has been updated since Nov, 2008. The world's moving fast, and I don't know if LVS was obsolete or not.  Is LVS standing still, or there're some better solutions to replace it?  Thank you so much. How to draw density estimate on histogram by pgfplots? I know how to draw a density histogram using pgfplots from this answer: .  But how can I draw an estimate line on the histogram like this?    \documentclass{article} \usepackage{pgfplots} \begin{document} \pgfplotsset{     small,     every axis plot post/.style={         fill=orange!75,         draw=orange!50!black     },     trim axis left }  \begin{tikzpicture} \begin{axis}[ymin=0, title=\texttt{hist=density}] \addplot [hist=density] table [y index=0] {random.txt}; \end{axis} \end{tikzpicture} \end{document} Erdős-Straus conjecture The  (ESC), states that for every natural number $n \geq 2$, there exists a set of natural numbers $a, b, c$ , such that the following equation is satisfied:  $$\frac{4}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\tag{1}$$  The basic approach to solving this problem outlined by Mordell [Ref1] is described below  By defining $t$ and $m$ as positive integers greater than zero and $q$ a positive integer greater than one we can observe that  a) There is always a solution for even $n$, since if $n=2^qt$ we have the trivial solution $$\frac{4}{4t}=\frac{1}{t}$$  In the remaining case $n=2(2t+1)$, a solution in the form of two Egyptian fractions can always be found  e.g. $$\frac{4}{2(2t+1)}=\frac{2}{2t+1}=\frac{1}{t+1}+\frac{1}{(t+1)(2t+1)}$$  b) If $(1)$ is a solution for some particular prime $n$ then all composite numbers $mn$ divisible by $n$ are also solutions, thus  $$\frac{4}{mn}=\frac{1}{ma}+\frac{1}{mb}+\frac{1}{mc}$$  will also be a solution. This means that we can simplify the analysis to the cases where $n$ is a prime greater than 2.  Using Mordell's approach we have just shown that we only need to consider the cases where $n$ is prime and where $n \equiv 1 \pmod{2} \;\;[meaning \;\;n=2t+1]$  The argument continues...  Mordell goes on to show in turn that the search can be reduced further to the cases when $$n \equiv 1 \pmod{4} \;\;[meaning \;numbers \;\;n=4t+1]$$ $$n \equiv 1 \pmod{8} \;\;[meaning \;numbers \;\;n=8t+1]$$ $$n \equiv 1 \pmod{3} \;\;[meaning \;numbers \;\;n=3t+1]$$ $$n \equiv 1,2,4 \pmod{7} \;\;[meaning \;numbers \;\;n=7t+1,n=7t+2 \;or\;n=7t+4 ]$$ $$n \equiv 1,4 \pmod{5}  \;\;[meaning \;numbers \;\; n=5t+1 \;or\;n=5t+4]$$  Assembling these results together, Mordell showed that the conjecture can be proved in this context except for the cases when $$n \equiv 1,11^2,13^2,17^2,19^2,23^2 \pmod{840}$$  Mordell stated that since the first prime meeting this condition is 1009, this is proof that the conjecture holds for $n&lt;1009$.  This basic approach can be pursued further. Other workers have shown that the conjecture holds for much higher values of $n$ using similar methods as can be seen on the above Wikipedia page.   Note that other intermediate results can be constructed from the above congruence's, e.g. $n \equiv 1 \pmod{24}$.  The question is:  Are there any other elementary approaches to solving this problem than the one outlined by Mordell (and described above)?  [Ref1] Louis J. Mordell (1969) Diophantine Equations, Academic Press, London, pp. 287-290. How to ask for "Save Changes?" on Lighthing Component form when user tries to navigate away We have a Lightning Community and would like to prompt the user to "Save Changes" if they try to click one of the community's navigational components (home link, or other menu item) but they have unsaved form changes on a custom Lightning component.  Ideally, we'd have a window.confirm popup asking if they want to leave without saving their changes, and if they click Cancel, the URL navigation would be aborted.  Is this possible via custom Lighting Components and Lighting Communities, and if so, how? If not, are there any patterns to work around this? Anyone know of a custom GPS ROM for Samsung Galaxy 5? Possible Duplicate:          I recently bought a Samsung Galaxy 5 (model number GT-I5500) and I'm new to the android world.   I'm not satisfied with the performance of the GPS on my cell phone so I thought it would be interesting to use a custom ROM to see if it improves a bit.   Anyone knows of a ROM that was tested in this model? prove that $\mathbb{Q}^n$is dense subset of $\mathbb{R^n}$ Proof of $\int_{[0,\infty)}pt^{p-1}\mu(\{x:|f(x)|\geq t\})d\mu(t)=\int_{[0,\infty)}\mu(\{x:|f(x)|^p\geq s\})d\mu(s)$ Let $({\Bbb R},{\mathcal A},\mu)$ be the measure space where ${\mu}$ is the Lebesgue measure.  Assume that $\int_{\Bbb R}|f|^pd\mu&lt;\infty$ ($p\geq1$). There is an exercise for proving that       $$\int_{[0,\infty)}pt^{p-1}\mu(\{x:|f(x)|\geq t\})d\mu(t)=\int_{[0,\infty)}\mu(\{x:|f(x)|^p\geq s\})d\mu(s).$$   If one formally let $s=t^p$, then $ds=pt^{p-1}dt$ and one would immediately get the equality above. However, how can I justify it rigorously in the sense of Lebesgue integration?    [Edited] What puzzles me is that I have the "change of variable" theorem for Riemann integration, but all the integrations here are in the Lebesgue sense. Unless one has shown that the integrands are also Riemann integrable, how can I directly use "change of variable"?