Is there an Android XML reference?
Is there any Android XML documentation?
Can I see a list of gyms I left a pokemon at? I captured a gym today just before the servers went down. I am wondering if I lost my progress. Is there a way to see if I own a gym or not? I develop a windows application and am curious whether it works via Wine. The application runs fine with Wine but I get no sound. In fact, I'm not sure whether my ubuntu installation has MIDI capabilities. The sound output settings shows "LFE on Separate Mono Output" one with "no amplifier" and another with "amplifier". They both function. I am using Ubuntu 12.04 as a client on virtualbox that runs on windows 7.  Thanks for any suggestion!  EDIT: virtual box hosted by windows 7, ubuntu running as a client in virtualbox. Does going outside the range of the spell or breaking line of sight between the caster and target break concentration or otherwise affect control over the spell?  For example, a Heat Metal spell has a range of 60 ft. If I cast the Heat Metal spell on a an armored foe, and that foe runs outside the range of the spell (60 ft), or around the corner, does that break concentration and end the spell? Heat Metal also allows you to use a bonus action to cause the 2d8 of damage — would exercising that option be prevented by being out of range or line of sight? I've been looking for some addon or something to render only the marked frames on timeline, i found this answer here on the stackexchange   The thing is this script only works with multiple cameras, i have only one camera that goes all aorund the scene but i only want to render the frames that i've marked on timeline.  Is there someone who can help me modify this script? Or find a better solution? Inverse of a matrix is expressible as a polynomial? Let $A$ be an $n \times n$ matrix. Prove that if A is invertible, then there exists a polynomial $p$, such that $A^{-1}=p(A)$  Thus far:  Let $W$ denote the $k$ dimensional A-cyclic subspace spanned by a vector $v$.  Then, $I_n=\sum_{i=0}^{k} a_{i}A^i$ for some scalar $a_i$.  What I want to do is multiply both sides by $A^{-1}$, so that way it just falls into my hands, but Im not sure that is 'legal'. I have a custom object. I want to put in a business rule so that a combination of two fields must be unique.   So I have Field1 and Field2 and want the combination to be unique:  Field1=4, Field2=7 Field1=4, Field2=8   is ok  But,   Field1=4, Field2=7 Field1=4, Field2=7   is not ok  What is the easiest / best way to do this? Thanks. Why can't I tame my ocelot? Irreducibility check for polynomials not satisfying Eisenstein Criterion. My new laptop came preinstalled with Windows 8. Naively, as I am, I just formatted the harddrive and installed fine old Ubuntu. Now I want to install Windows 8 for dual boot again, but I have no DVD and do download the ISO one needs a product key. That key is not on the back of the laptop anymore but somewhere on the mainboard.  Is there any way to recover the product key from the mainboard using Ubuntu? if $f^{(n)}(x_0),g^{(n)}(x_0)$ exist and $\lim_{x\to x_0} \dfrac{f(x)-g(x)}{(x-x_0)^n}=0$ then $f^{(r)}(x_0)=g^{(r)}(x_0 )$ for all $0\leq r\leq n.$ Show:   if $f^{(n)}(x_0),g^{(n)}(x_0)$ exist and   $$\lim_{x\to x_0} \dfrac{f(x)-g(x)}{(x-x_0)^n}=0$$   then $f^{(r)}(x_0)=g^{(r)}(x_0 )$ for all $0\leq r\leq n.$   My attempt:  let n = 1,  since $f'(x_0),g'(x_0)$ exists then The taylor polynomial exists: $$T_{f,1}(x)=f(x_0)+f'(x_0)(x-x_0), \lim_{x\to x_0} \dfrac{f(x)-T_{f,1}(x)}{(x-x_0)}=0$$  $$T_{g,1}(x)=g(x_0)+g'(x_0)(x-x_0), \lim_{x\to x_0} \dfrac{g(x)-T_{g,1}(x)}{(x-x_0)}=0$$  Then, \begin{align} \lim_{x\to x_0} \dfrac{f(x)-g(x)}{(x-x_0)^n} &amp; =\lim_{x\to x_0} \dfrac{f(x)-g(x)+T_{f,1}(x)-T_{f,1}(x)+T_{g,1}(x)-T_{g,1}(x)}{(x-x_0)^n}\\ &amp;=\lim_{x\to x_0} \dfrac{f(x)-g(x)+T_{f,1}(x)-T_{f,1}(x)+T_{g,1}(x)-T_{g,1}(x)}{(x-x_0)^n}\\ &amp;=\lim_{x\to x_0} \dfrac{f(x)-T_{f,1}(x)}{x-x_0}-\lim_{x\to x_0} \dfrac{g(x)-T_{g,1}(x)}{x-x_0}+\lim_{x\to x_0} \dfrac{T_{f,1}(x)-T_{g,1}(x)}{x-x_0}\\ &amp;=\lim_{x\to x_0}\dfrac{T_{f,1}(x)-T_{g,1}(x)}{x-x_0}\\ &amp;=0 \end{align}  Then, \begin{align} \lim_{x\to x_0}\dfrac{T_{f,1}(x)-T_{g,1}(x)}{x-x_0} &amp;=\lim_{x\to x_0}\dfrac{f(x_0)+f'(x_0)(x-x_0)-g(x_0)-g'(x_0)(x-x_0)}{x-x_0}\\ &amp;=\left(f(x_0)-g(x_0)\right)\lim_{x\to x_0}\dfrac{1}{x-x_0}+f'(x_0)-g'(x_0) \\&amp;=0 \end{align} I am confused because  $\lim_{x\to x_0}\dfrac{1}{x-x_0}$ does not exist. Is it possible to get the sheetname? I know there's some JAVA_OPTS to set to remotely debug a Java program.  What are they and what do they mean ? Find all solutions to $x^2\equiv 1\pmod {91},\ 91 = 7\cdot 13$ The problem is stated as follows:  "Let $R$ be a Noetherian ring and $\theta$ be a ring homomorphism from $R$ to $R$. Show that if $\theta$ is surjective then it is also injective."  Regardless of the right solution, I don't understand why is the following wrong:     We have $\theta: R\to R$. By the isomorphism theorem $R/\ker\theta\cong\operatorname{Im}\theta$. Since $\operatorname{Im}\theta = R$, it follows $\ker\theta =0$, so it's injective.   Harsh criticism will be appreciated. Thanks. When to use 'which' or 'in which' or 'that' (as relative pronouns)? Is complex conjugation needed for valid inner product? What are the benefits of using a conjugate linear inner product in a complex vector space vs a simple linear inner product?  That is, why do we demand that $(y,x) = \overline{(x,y)}$ as opposed to $(y,x)=(x,y)$?  Of course, this ensures that $(x,x)$ is real and thus makes an easy definition of norm, but is that necessary? Understanding the proof for showing that there are infinitely many $n\in\mathbb{Z}^+$ such that $n!+1$ is divisible by at least two distinct primes There is another post pertaining to this question; however, the other post does not address some specifics that concern me in the proof. Proof: Let $n=p-1, p \geq 7$. Then $p|(p-1)!+1$. We will show that $(p-1)!+1=p^n$ has no solutions. Suppose not. Suppose $(p-1)! + 1 = p^n$ has a solution for $p\geq 7$. Then $(p-1)! =p^n-1  = (p-1)(p^{n-1}+\cdots+p+1)$. So $(p-2)!\equiv n (\mod p-1)$. Now suppose that $(p-1)|n$. Then $p-1\leq n$. So $(p-1)!+1&lt;p^{p-1}\leq p^n$. Hence, there are no solutions. It remains to show that $(p-1)|(p-2)!$. But $p-1=2m$ means $2m|(p-2)!$. QED  Questions:  My understanding is that at the start of the proof, Wilson's theorem is used to show that at least one prime divides $n!+1$.  Why do we want to show that $(p-1)!+1=p^n$ has no solutions? The way I see it is that $p^n$ is composite, and we are basically saying that $p^n\nmid (p-1)!+1$, so no composite number divides $n! +1$. But how and where in the proof do we actually show that another prime $q \neq p$ divides $n!+1$?    Thanks in advance. Difference estimate & confidence intervals for $\chi^2$ test between 2 proportions I am using a chi-sq test between two proportions (). I am using this over the z-test for proportions because I do not think my data is normal.   My question is how do I get the difference (shift) estimate &amp; confidence intervals for this test.   For example,     Data1: 193/252=.77   Data2: 154/227=.68   P-value=.032 (I used the formula in the link above. I will also add my exact code below)  Shift Estimate: I assume its the difference between proportions, ie .77-.68=.09. Is this correct?   Confidence Interval: ??????????????????  Code for generating P-value (matlab):   % Pooled estimate of proportion  p0 = (n1+n2) / (N1+N2); % Expected counts under H0 (null hypothesis)  n10 = N1 * p0;  n20 = N2 * p0; % Chi-square test, by hand  observed = [n1 N1-n1 n2 N2-n2];  expected = [n10 N1-n10 n20 N2-n20];  chi2stat = sum((observed-expected).^2 ./ expected);  p = 1 - chi2cdf(chi2stat,1);  H=0; if(p&lt;.05), H=1; end Showing that $|f(z)| \leq \prod \limits_{k=1}^n \left|\frac{z-z_k}{1-\overline{z_k}z} \right|$ Which predictions of The Last Jedi: Darths & Droids Bingo were correct?