Update multiple Windows 8 to 8.1
Is there a way I can share the 3.42 GB download of Windows 8.1 from Windows Store?
Something weird is going on with my login on at least one site If $T^n$ is $q$-contractive, $T$ exactly has one fixed point Consider a complete metric space $(X,d)$ and $T\colon X\to X$. Suppose there exists $n\in\mathbb{N}$ such that the n-th power of $T$ is $q$-contractive. Show that then $T$ has exactly one fixed point $\overline{x}\in X$.      The n-th power is defined inductively: $$ T^{n+1}(x):=T(T^{n}(x)), n\in\mathbb{N}. $$ And for a function $T\colon X\to X$, $q$-contractive means  $$ \exists 0\leq q&lt;1~\forall~x,y\in X: d(T(x),T(y))\leq q d(x,y). $$    Now to the proof.  I think I have to show, that $T$ is q-contractive, because then it follows with Banach, that $T$ does have exactly one fixed point. So I have to show, that there exists a $0\leq q&lt;1$ so that for all $x,y\in X$ it is $$ d(T(x),T(y))\leq q\cdot d(x,y). $$ And most likely I have to use the q-contractivity of $T^n$, i.e. that there exists a $0\leq q &lt;1$ so that for all $x,y\in X$ it is $$ d(T^n(x),T^n(y))\leq q\cdot d(x,y). $$  Can you help me? I've begun cooking Asian food in the past year or so, and in many Korean and Chinese beef recipes, I see instructions to soak beef in water. Just plain water, not salt water or anything. And every recipe seems to have a different  time - from 5 minutes to like, multiple hours.  I inquired about this and was given various reasons for this step, including "to remove the blood (because it is simply undesirable", "To remove the blood, because it gives the meat a bad taste" or "To soften the beef".  I'm wondering:   Does soaking meat in water really remove that much blood? How much of a flavor component is blood and why would it give it a bad taste (especially since i've never soaked beef before and I've like the taste just fine)? What is the action of solute-free water on a piece of beef immersed in it? Can it really tenderise it to any degree, or is there any osmosis occurring or other such physical/chemical action? What would be the value of $\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}$ I would like to evaluate the sum  $$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$  Here is my attempt:    Letting    $$f(z)=\frac{1}{az^2+bz+c}$$  The poles of $f(z)$ are located at   $$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$  and  $$z_1 = \frac{-b-\sqrt{b^2-4ac}}{2a}$$  Then  $$ b_0=\operatorname*{Res}_{z=z_0}\,\pi \cot (\pi z)f(z)= \lim_{z \to z_0} \frac{(z-z_0)\pi\cot (\pi z)}{az^2+bz+c}= \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b} $$  Using L'Hopital's rule. Continuing, we have the limit is  $$ \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}= \frac{\pi\cot (\pi z_0)}{2az_0+b} $$  For $z_0 \ne 0$  Similarly, we find  $$b_1=\operatorname*{Res}_{z=z_1}\,\pi \cot (\pi z)f(z)=\frac{\pi\cot (\pi z_1)}{2az_1+b}$$  Then   $$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c} = -(b_0+b_1)=\\ -\pi\left( \frac{\cot (\pi z_0)}{2az_0+b} + \frac{\cot (\pi z_1)}{2az_1+b}\right)=  -\pi\left( \frac{\cot (\pi z_0)}{\sqrt{b^2-4ac}} + \frac{\cot (\pi z_1)}{-\sqrt{b^2-4ac}}\right)= \frac{-\pi(\cot (\pi z_0)-\cot (\pi z_1))}{\sqrt{b^2-4ac}}= \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{\sqrt{b^2-4ac}} $$  Then we have  $$\sum_{n=0}^\infty \frac{1}{an^2+bn+c} = \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{2\sqrt{b^2-4ac}}$$    Is this correct?  I feel like I made a mistake somewhere.  Could someone correct me?  Is there an easier way to evaluate this sum? Solving 4 Quadratic Simultaneous Equations Consider the equations:   $$\begin{cases} ab+c+d=3\\ bc+d+a=5\\ cd+a+b=2\\ da+c+b=6 \end{cases}$$   Where $a,b,c,d \in \mathbb{R}$. How can we find $a,b,c,d$?   The furthest I've got is by adding the first two equations and the last two equations together to get: $ab+bc+c+d+d+a=(b+1)(a+c)+2d=8$ and $(d+1)(a+c)+2b=8$ We can rearrange these (assuming $b,d \neq -1$) to get $$\frac{8-2d}{b+1}=\frac{8-2b}{d+1} \implies -d^{2}+3d+4=-b^{2}+3b+4$$ Which leads us further to $$b^{2}-d^{2}-3(b-d)=0 \implies (b-d)(b+d-3)=0$$ Now since $b=d$ gives us an absurdity (equations 1 and 4 reduce to 3=6), we must have $b+d=3$. This is as far as I can get unfortunately, since trying the same approach with the other pairs doesn't work in the same way.  Help would be much appreciated; the problem is from the BMO1 2004 paper. Shouldn't it be "he was the most famous fighter to have ever lived"? How to list files installed by a snap package? After a snap has been installed with sudo snap install [package] how can I list the files which have been installed by the snap? I told my wolves to sit down and I accidentally forgot to tell them to stand up and I left them. Now I cannot find them, how do you get back to your wolves if they are lost? I am flying to Tbilisi, Georgia with Turkish Airline. My stay in Istanbul is 12 hours. Can I get transit visa to visit Istanbul city ?    I am from Pakistan with Oman employment visa. How can I combine hashes in Perl? What is the best way to combine both hashes into %hash1?  I always know that %hash2 and %hash1 always have unique keys.  I would also prefer a single line of code if possible.    $hash1{'1'} = 'red'; $hash1{'2'} = 'blue'; $hash2{'3'} = 'green'; $hash2{'4'} = 'yellow'; CPT Template Not Showing - Getting 404 I'm developing a Custom Post Type plugin, which works fine in dev. Now that I've moved it to production, I'm getting a 404 when I try to display the template for the CPT.   So, two questions:  1- Is there some mechanism to determine the path WP took to get through the hierarchy?  2- Why wouldn't it just give me the single or index page?  I've dumped $wp_query and a stacktrace at the end of the 404. Here's part of what I found:  [query] =&gt; Array     (         [page] =&gt;          [pagename] =&gt; refletters/proximity     )  [request] =&gt; SELECT   wp_posts.* FROM wp_posts  WHERE 1=1  AND (wp_posts.ID = '0') AND wp_posts.post_type = 'page'  ORDER BY wp_posts.post_date DESC  [posts] =&gt; Array     (     )  [queried_object] =&gt;  [queried_object_id] =&gt; 0   And the stack trace:  2013-04-27 22:16:41 - 24.52.197.40 - TRACE - index.php - 404.php[39] -      wordpress/wp-content/themes/notoriousconsultant/404.php[39] - aaLog-&gt;logtrace     wordpress/wp-includes/template-loader.php[50] - include     wordpress/wp-blog-header.php[16] - require_once     wordpress/index.php[17] - require   Thanks for your help. There are many collectable flags throughout the game of Assassin's Creed and every time one is collected, you get a progress indicator (e.g., 33 of 100).  Is there any reason to collect these, other than bragging rights?  I recall getting all available flags in one particular town and never saw any special reward or secret revealed. Bound on the difference of two determinants Let $A$ and $B$ be two real, $n\times n$ matrices. Using Hadamard's inequality, it is not hard to show that $$ \left|\det A - \det B \right| \leq \|A-B\|_{2} \frac{\|A\|_{2}^n -\|B\|_{2}^n}{\|A\|_2 -\|B\|_2}. $$ Where $\|A\|_2=\sqrt{\sum_{i,j}a_{ij}^2}$.  From this, I can derive a sup bound, for example  $$ \left|\det A - \det B \right| \leq n^{n+1} \|A-B\|_{\infty} \max (\|A\|_{\infty}^{n-1},\|B\|_{\infty}^{n-1}). $$ Where $\|A\|_\infty=\sup_{i,j}|a_{ij}|$.   The constant $n^{n+1}$ is not the best bound possible : any reference (or proof) for a better (or the best) one? I show below that one can obtain $n^2(n-1)^{n-1}$, but that isn't much better. I just tried  $10^5$ random matrices on Maple and obtained a maximal constant (much) smaller  than one : this is not a proof, but it looks like there is room for improvement nevertheless.    Just for completeness (and in case someone sees a factor I missed), to get the first bound, writing $A=[A_1,\ldots,A_n]$ in terms of its column vectors, an expansion shows \begin{eqnarray*} \det A &amp;=&amp; \det (A_1 -B_1,A_2,\ldots,A_n) + \det (B_1,A_2,\ldots,A_n) \\   &amp;=&amp; \sum_{j=1}^n \det (B_1,\ldots, B_{j-1}, A_j -B_j,A_{j+1},\ldots,A_n) \\ &amp;&amp; + \det B,  \end{eqnarray*} Thus by Hadamard's inequality, \begin{eqnarray*} \det A -\det B &amp;\leq&amp;  \sum_{j=1}^n \prod_{i=1}^{j-1} \|B_i\|_{2}\prod_{i=j+1}^{n} \|A_i\|_{2} \|A_j-B_j\|_{2} \\ &amp;\leq&amp;  \|A-B\|_{2} \sum_{j=1}^n \|B\|^{j-1}_{2}\|A\|^{n-j}_{2} \\  &amp;=&amp; \|A-B\|_{2} \frac{\|A\|_{2}^n -\|B\|_{2}^n}{\|A\|_2 -\|B\|_2}. \end{eqnarray*}  The second bound is just that $x^n -y^n\leq n \max(|x|^{n-1},|y|^{n-1}) |x-y|$ and $\|A\|_2 \leq n\|A\|_\infty$.    Another approach is calculus, namely, to write that $\det B - \det A = f(1)-f(0)$, with $f(t)=\det(A + t(B-A))$.   By the mean value theorem $f(1)-f(0)\leq \max |f^\prime (t)|$.   We can compute that $$f^\prime(t) =  {\rm trace}\left({\rm Cofm}(A +t(B-A))(B-A)\right)$$  (if I did not mess up, using the formula for the differential of a determinant, where Cofm means the matrix of Cofactors).   Then, it should deliver something better, if there is a nice way to bound it. The simplest thing is to use Cauchy-Schwarz, namely $$ \left|{\rm trace}\left({\rm Cofm}(A +t(B-A))(B-A)\right)\right|\leq \|B-A\|_{2} \|{\rm Cofm}(A +t(B-A))\|_{2} $$ and then, by lack of a better idea,  $$ \|{\rm Cofm}(A +t(B-A))\|_{2}\leq n \max_{ij} |{\rm Cof}_{i,j}(A +t(B-A))|, $$ and brutally, $|{\rm Cof}_{i,j}(A +t(B-A))|\leq ((n-1) \max(\|A\|_\infty,\|B\|_\infty))^{n-1}$ gives a slightly better constant, namely $$ n^2(n-1)^{n-1} &lt;n^{n+1} $$ but that still seems a very rough way to bound a determinant, as it is never sharp, since to attain this bound all coefficients should be equal, and therefore the cofactor would be zero. They are of the same order in $n$, and I suspect this order is wrong. How to create animated GIF images of a screencast? I've seen animated GIF images of screen casts (like the one below) promoted a few times on this site as a way to improve answers.    What toolchain is being used to create these? Is there a program that does this automagically, or are people taking screencasts, converting them into a series of static frames, and then creating the GIF images? How to save Django ModelForm data? Prevent Windows 10 from installing/updating a specific driver? It seems that the recent Synaptics driver update for Synaptics SMBus TouchPad on Windows 10 (build 100049) somehow doesn't see my middle-button. Uninstalling, and installing an older driver for Windows 8.1 from the HP site (my machine is HP EliteBook 8570w) restores the middle button functionality.    Few days later, the driver updates itself again, and I lose the middle-button.  I realize I'm running unsupported software, but is it possible to prevent updating this particular driver, without either disabling updates altogether or preventing other drivers from updating? Why are p-values uniformly distributed under the null hypothesis? Showing that $\lnot Q \lor (\lnot Q \land R) = \lnot Q$ without a truth table I've done a truth table after reducing it to this and it seems to be equal to $\neg Q$:  $$\lnot Q \lor (\lnot Q \land R) = \lnot Q$$  But when I try to show it without a truth table (with just transformations), I end up in a loop between that and:  $$\lnot Q \land (\lnot Q \lor R)$$  Is there a way to show this is true without using a truth table? What am I missing?!  Thanks in advance! Why should I declare a virtual destructor for an abstract class in C++? I know it is a good practice to declare virtual destructors for base classes in C++, but is it always important to declare virtual destructors even for abstract classes that function as interfaces? Please provide some reasons and examples why. No group of order 400 is simple