I have installed CyanogenMod 10.1.x on my Note 2, but I find it very unstable and it often goes on a reboot loop.  I use ClockworkMod recovery.  I would rather work with a stock android at this point.  Questions:   Since I have ClockworkMod recovery, I presume I can use it to flash the "official" ROM? Is this correct? Can someone point me to the official stock android from Samsung?
There are several reasons to be looking for a ROM:   Upgrading or downgrading your stock ROM Looking for firmware from a different region/language Flashing a custom ROM for different features, removing bloat, etc.   But where can one find ROMs in the first place?  What sites are likely to have the ROM I need?  ; please add to the wiki answer instead of posting separate answers.  See also:
What is the maximum number of people that will join a settlement? I've got two settlements on the go that I'm paying attention to - Sanctuary and the Starlight Drive In. Both of these settlements have radio towers and beds and all resources in the green however they appear to have stopped growing at 10 people in each settlement even though happiness is above 80 and I've got 15+ beds at each location.  What is the maximum number of people that will join a settlement? How can one differentiate between Dr. (PhD) and Dr. (MD or DO)? This is the problem I am currently having trouble with:  If $\lambda_1$ and $\lambda_2$ are eigenvalues of matrix  $$         \begin{bmatrix}         a &amp; b\\         c &amp; d\\         \end{bmatrix}$$prove that $\lambda_1^2$, $\lambda_1\lambda_2$ and $\lambda_2^2$ are eigenvalues of matrix $A$.     $$A=\begin{bmatrix}     a^2 &amp; ab &amp; b^2\\     2ac &amp; ad+bc &amp;2bd\\     c^2 &amp; cd &amp;d^2\\     \end{bmatrix}$$  What have I considered so far:   I tried to find characteristic polynomial of first matrix and then find the roots which would be $\lambda_1$ and $\lambda_2$. I was hoping I can make a connection with characteristic polynomial of matrix $A$, but it got too complicated and did not seem like the right idea. Since there are two eigenvalues for first matrix, there are two eigenvectors. I thought about somehow using $A\vec v_i=\lambda_i\vec v_i$, where $i=1,2$  Also, I thought about finding a connection between these two matrices. What I see is that the sum of elements of first and second column, of first matrix, are $(a+c)$ and $(b+d)$, respectively.  Sum of first, second and third column of matrix $A$ are $(a+c)^2, (a+c)(b+d)$ and $(b+d)^2$. Now, if I could assign $\lambda_1$ and $\lambda_2$ to $(a+c)$ and $(b+d)$, then I would get that sum of first, second and third column of matrix $A$ are $\lambda_1^2$, $\lambda_1\lambda_2$ and $\lambda_2^2$, but what would be next?   I thought it would be good to combine 2. and 3rd idea, but I am not making any progress.  I appreciate any help. How do I invoke a private static method using reflection (Java)? VMWare: This host doesn't support VT. Can not play 64-bit VM due to lack of VT flag I have an Intel E7300 Core2 Duo processor with 2.66GHz speed. I need to run a 64-bit WinXP SP2 VMWare image. My host operating system is Win 7. So I just installed the VMPlayer from www.vmware.com.  The problem is when I run the VM image it says This host doesn't support VT.  If I continue I see Windows shows an error message saying     Attempting to load an x64 operating system, however this CPU is not compatible with x64 mode.   My question is how do I run this VM without buying a new processor? Any other tools or software? Should we write “allow” or ”allows” in “require that the class allow[s]”? Can you explain to me whether I should or should not use ‑s at the end of the  verb allow in this sentence?     This function requires that the container class allow[s] random access   The sentence in the book I read wrote allow without ‑s. However, the container class is a third-person singular, so the ‑s must be required. Is that right? What is the Page Object Pattern in Selenium WebDriver? Do I make money in the stock market from other people losing money? Clicking on tags in global search results is broken If you search for something on "all sites" (using the searchbox at the top of , you'll reach the search results, e.g.  .  If you click directly on a tag on that page, e.g.        You get 404 Page Not Found error, e.g.     Steps to reproduce   Go to  and enter e.g. "keyforge" into the "Search all sites" field. Arrive at  Click on the "keyforge" tag below a search entry. Reach  (404 page not found)   Expected behaviour  Reach   System Info  Happens in both Firefox and Vivaldi browsers on Windows 10. How can I skip every even item in the enumerate environment without having to continually type \setcounter{enumi}{5}? I am still learning to program macros, so I spologize if this is a ridiculously easy question.   Just to be clear, I would like something like   1. *Something.* 3. *Somthing else.* 5. *A third something.*  ...As opposed to   1. Something.  2. A second something.  3. A tertiary Something. let q be an odd prime and p a prime dividing $1+n+\dots+n^{q-1}$. Prove that $p = 1 \mod q$ or $p = q$.  I don't actually know how to start, i thought of using fermat's little theorem or an application of eulers formula, but it didn't really give me any insights.   Any hints? Let $G$ be a non-empty finite set with an associative binary operation so that cancellation law holds, i.e. $ab=ac$ or $ba=ca$ implies $b=c$, for any choices of $a,b,c$ in $G$. Assume that there is an identity element $e$ in $G$. Show that $G$ is a group.  Proof: To show $G$ is a group, conditions must hold. Suppose that $ab=ac$, then $b=eb=(a^{-1}a)b =a^{-1}(ab)=a^{-1}(ac) =(a^{-1}a)c=ec=c$. So left cancellation holds in $G$.  2) Suppose $ba=ca$, then $b=be=b(aa^{-1}) =(ba)a^{-1}=(ca)a^{-1} =c(aa^{-1})=ce=c$. So right cancellation holds in $G$.  3) If $a$ exists in $G$ then $aa^{-1}=a^{-1}a=e$, where $a$ is an inverse of $a^{-1}$. Since inverses are unique, $(a^{-1})^{-1}=a$.  4) Let $x$ be the inverse of $ab$. Then $(ab)x=e$. By associativity we have $a(bx)=aa^{-1}$. Through left cancellation we have  $bx=a^{-1}bx=ea^{-1}=b(b^{-1}a^{-1})$ and $x=b^{-1}a^{-1}$.  Thus $(ab)^{-1}=b^{-1}a^{-1}$. So all conditions hold, $G$ is a group.  Is this proof correct? I know to show $G$ is a group these conditions have to be met. This is all I have to show right? Windows 7 stuck on CLASSPNP.SYS I'm running Win 7 x86 on a Lenovo Thinkpad Z61m in dual boot mode with Win XP, both OSes are running on separate partitions. I have a faulty battery and if left unplugged my laptop shuts usually down within 10 minutes. Last night, it shut down due to this, but after plugging in AC power and starting back up I'm stuck at the "Starting Windows" screen, I'm unable to boot into safe mode or into recovery mode(XP boots fine).  I believe I had installed some Win 7 updates the same/previous day. From one of the boot options I was able to see that Windows 7 is getting stuck on CLASSPNP.SYS  How can I go about fixing this?  I do have a complete Win 7 Restore CD that I created after setting up my machine, I'm unable to boot into this either.  Please help! ERROR running force:source:pull: Entity of type 'CustomApplication' named 'Contact' cannot be found I am looking for a closed form (ideally expressed as elementary functions) of the function $\exp_n(x) = \sum_{k=0}^n x^k / k!$.  I am already aware of expressing it in terms of the gamma function. Background / Motivation When counting combinations of objects with generating functions, it is useful to be able to express the partial sum $1 + x + \cdots + x^n$ as $\frac{1-x^{n+1}}{1-x}$. For example, to count the number of ways to pick 5 marbles from a bag of blue, red, and green marbles where we pick at most 3 blue marbles and at most 2 red marbles, we can consider the generating function $f(x) = (1+x+x^2+x^3)(1+x+x^2)(1+x+x^2+\cdots)$. By using the partial sum identity, we can express it as $f(x) = \left(\frac{1-x^4}{1-x}\right)\left(\frac{1-x^3}{1-x}\right)\left(\frac{1}{1-x}\right)$.  Simplify, express as simpler product of series, and find the coefficient of the $x^5$ term. I want to be able to do the same for a generating function in the form $g(x) = \exp_{n_1}(x)^{p_1} \exp_{n_2}(x)^{p_2} \cdots \exp_{n_j}(x)^{p_j}$ The easiest way to extract the coefficient of a given term $x^p / p!$ would be to use a similar closed form expression for $\exp_n(x)$ and a similar technique to $f$. Attempted Solutions Differential equation Recall that the way to prove the identity $1+x+x^2+\cdots+x^n = \frac{1-x^{n+1}}{1-x}$ is to define $S = 1 + x + x^2 + \cdots + x^n$ and notice that: $S - Sx = 1 - x^{n+1}$. Likewise, notice that $y(x) = \exp_n(x)$ satisfies $y - y' = x^n/n!$.  Via SAGE, the solution is $y(x) = \frac{c+\Gamma(n+1,x)}{n!}e^x$.  Our initial condition $y(0) = 0$ so $c=0$.  By , $\Gamma(n+1,x) = n! e^{-x} \exp_n(x)$ so we have $y(x) = \exp_n(x)$. Recurrence Relation Notice that $\exp_n(x) = \exp_{n-1}(x) + x^n/n!$. Using the unilateral Z-Transform and related properties, we find that $\mathcal{Z}[\exp_n(x)] = (z e^{x/z})/(z-1)$. Therefore, $\exp_n(x) = \mathcal{Z}^{-1}\left[(z e^{x/z})/(z-1)\right] = \frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz$. $(z^n e^{x/z})/(z-1)$ has two singularities: $z = 1$ and $z = 0$.  The point $z = 1$ is a pole of order one with residue $e^x$.  To find the residue at $z = 0$ consider the product $z^n e^{x/z} (-1/(1-z)) = -z^n \left( \sum_{m=0}^\infty x^m z^{-m} / m! \right) \left( \sum_{j=0}^\infty x^j \right)$.  The coefficient of the $z^{-1}$ term is given when $n - m + j = -1$. The residue of the point $z=0$ is then $-\sum_{m,j} x^m / m! = -\sum_{m=n+1}^\infty x^m / m!$. Let $C$ by the positively oriented unit circle centered at the origin.  By Cauchy's Residue Theorem, $\frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz = \frac{1}{2 \pi i} 2 \pi i \left(e^x - \sum_{m=n+1}^\infty x^m / m!\right) = \exp_n(x)$. Finite Calculus I've tried to evaluate the sum using finite calculus, but can't seem to make much progress. The equation $X^{n} + Y^{n} = Z^{n}$ , where $ n \geq 3$ is a natural number, has no solutions at all where $X,Y,Z$ are intergers. In QGIS I would like to clip a raster precipitation layer using an admin boundary layer that is vector data. However the Geoprocessing tools seem to be usable only for vector data. How can I clip this precipitation layer? How to call shell commands from Ruby How do I call shell commands from inside of a Ruby program? How do I then get output from these commands back into Ruby? How do I change Wi-Fi Hotspot settings in Ubuntu with GNOME 3? How do I manage my hotspot connected devices or change the hotspot password? I can only see the Network Name, Security type and the Password but I can't seem to manage users or the network properties. Do you need to erase the memory of *watching* a memory in a Pensieve? You remove your memory from your brain to store in a Pensieve (as Snape or Dumbledore did). Then, you re-watch the memory (e.g. as Dumbledore re-watched his memories of Voldemort's investigations with Harry) in the Pensieve.   Does that mean that you now re-created the memory in your brain (because you remember watching it) and thus you'll need to extract/erase it from your brain again if you don't wish to keep it there for security/capacity reasons (which was why you stored it in the Pensieve in the first place?  Canon answers only please.