How do I fix vertices that aren't following the armature?
Weight Painting Won't Get Rid of Weird Protrusions
Backup to Azure Storage Container: Shared Access Signature the only way? RESTORE FILELISTONLY FROM URL = '' WITH CREDENTIAL = '' fails in SQL Server 2016.  Copied script from SQL Server 2012 environment, successfully restoring from Azure Storage Container, with create credential statement, and it failed in 2016.  Error message "Forbidden".  Only after logging into the Azure Portal, creating a Shared Access Signature for the container, could I execute the restore, and only by not specifying the credential did the restore work.  No documentation advises that a SAS (Shared Access Signature) is required in SQL Server 2016 for Azure container backup / restore.  Is there another way to use T-SQL to access blob backups on Azure storage containers in SQL Server 2016, than using SAS?    Looking at the MSDN tutorial on how to configure the backup service SQL 2016 introduces, a SAS is used. Refer External Styles in JasperReports I'm working on a system that includes a large number of reports, generated using . One of the newer features is that you can define styles for reports.  From the available docs I believe there is some way to have an external file defining styles to use, and you can reference that in your jasper reports. This allows a single style to be used by multiple reports.  I can't find any concrete information on whether this is an actual feature, and if it is, how to use it. Does anyone know if it is possible to have external styles for jasper reports, and if so, how to do it? Group of order $p^2$ is abelian. Yes, I know that there are tons of solutions of this up here, but I, essentially, wanted to try it a different way and ah, well.  Let $|G| = p^2$ for some prime $p$.  Consider $x \in G$. So, $|x| = p, $ or $ p^2$. If it is the latter, then, we are done.  So, if $|x|= p$, then, let, $H = [h|  h=x^i, i \in Z] \Rightarrow |H|= p$.  As $p$ is the smallest prime dividing $|G|$, $H$ is a normal subgroup.  Now, consider some $y \in G$, but not in $H \Rightarrow y^p= e$. Then, its easy to show that $G = \langle x,y \rangle$.  Now, consider a homomorphism $f:G \rightarrow \operatorname{Aut}(H), g \to c_g $, wherein $c_g$ represents conjugacy by $g$. So, for example, $c_g(x) = gx g^{-1}$.  Then, it is clear that, $H \subset \operatorname{Ker}(f) $, as it is cyclic and therefore abelian. So, we need only look at $G/H$.  Basically, I am asking for a hint as to why $c_y(x) = yxy^{-1} = x = c_e(x)$, where $x$ and $y$ are the generators of $H$ and $G/H$, respectively?  Because, if I establish it for the generators, then it follows that it applies for for all elements, i.e, that $\operatorname{Im}(f)$ is trivial and therefore that $G$ is abelian.  Thank You! I am wondering why matrix factorization techniques in the machine learning domain almost always expect the provided matrix to be non-negative. What is the advantage of this constraint?  Background: I want to use matrix factorization algorithms for a sparse user-item matrix containing positive and negative implicit feedback. Is there any another possibility to set interactions with negative indications apart from fields that denote that no interaction happened between the user and the item? BAD signature when trying to verify SHA256SUMS for Bionic Beaver ISO I am attempting to verify my download of Bionic Beaver on Windows 7.  I feel like I am missing something very obvious.  I went to  and clicked on both SHA256SUMS and SHA256SUMS.gpg  It seems like this does not download the files, but opens the text in the files in new tabs.  I used Ctrl-A and then copied the text into Notepad, naming the files SHA256SUMS and SHA256SUMS.gpg.  I made sure that windows didn't add any weird extensions when saving.  I manually downloaded both keys I used from https://keyserver.ubuntu.com, one with fingerprint  C598 6B4F 1257 FFA8 6632 CBA7 4618 1433 FBB7 5451   and the other with fingerprint  8439 38DF 228D 22F7 B374 2BC0 D94A A3F0 EFE2 1092   After I downloaded both keys I certified them with my personal key using Kleopatra, after verifying the key fingerprint with those at .  When I ran   gpg --verify SHA256SUMS.gpg SHA256SUMS   I got   gpg: Signature made 11/29/18 16:27:43 US Mountain Standard Time gpg:                using DSA key 46181433FBB75451 gpg: BAD signature from "Ubuntu CD Image Automatic Signing Key &lt;cdimage@ubuntu.c om&gt;" [full] gpg: Signature made 11/29/18 16:27:43 US Mountain Standard Time gpg:                using RSA key D94AA3F0EFE21092 gpg: BAD signature from "Ubuntu CD Image Automatic Signing Key (2012) &lt;cdimage@u buntu.com&gt;" [full]   When I ran   gpg --verbose --verify SHA256SUMS.gpg SHA256SUMS   I got  gpg: armor header: Version: GnuPG v1 gpg: Signature made 11/29/18 16:27:43 US Mountain Standard Time gpg:                using DSA key 46181433FBB75451 gpg: using pgp trust model gpg: BAD signature from "Ubuntu CD Image Automatic Signing Key &lt;cdimage@ubuntu.c om&gt;" [full] gpg: binary signature, digest algorithm SHA512, key algorithm dsa1024 gpg: Signature made 11/29/18 16:27:43 US Mountain Standard Time gpg:                using RSA key D94AA3F0EFE21092 gpg: BAD signature from "Ubuntu CD Image Automatic Signing Key (2012) &lt;cdimage@u buntu.com&gt;" [full] gpg: binary signature, digest algorithm SHA512, key algorithm rsa4096   I tried copying the data from SHA256SUMS and SHA256SUMS.gpg tabs over TOR instead, in case it was a local network problem, but no dice either.  What am I missing here? Is there some way to download the SHA256SUMS and SHA256SUMS.gpg files directly, as opposed to copying the data into a text editor and saving it?  Any help would be very appreciated. I am very confused and frustrated at this point. I read in the book Applied Analysis by Hunter and Nachtergale that     the sequence $x(n)=\log(n)$ is not Cauchy since $\log(n)\to\infty$   But that seems to be irrelevant to the definition of a Cauchy sequence which I understand is as follows:     A sequence $x(n)$ is said to be Cauchy if for every $\epsilon &gt; 0$  there exists an $N$ such that $\lvert x(m)-x(n)\rvert &lt; \epsilon$ for all $m,n&gt;N$.   This sequence $(\log n)$ seems to meet the definition. So how come it is not considered Cauchy? Are the eigenvalues of $AB$ equal to the eigenvalues of $BA$? (Citation needed!) First of all, am I being crazy in thinking that if $\lambda$ is an eigenvalue of $AB$, where $A$ and $B$ are both $N \times N$ matrices (not necessarily invertible), then $\lambda$ is also an eigenvalue of $BA$?  If it's not true, then under what conditions is it true or not true?  If it is true, can anyone point me to a citation?  I couldn't find it in a quick perusal of Horn &amp; Johnson.  I have seen a couple proofs that the characteristic polynomial of $AB$ is equal to the characteristic polynomial of $BA$, but none with any citations.  A trivial proof would be OK, but a citation is better. nameref (hyperref) evaluating counter instead of section name What's the proper way to reference a section with a counter in the title? The TOC generates correctly, but the cross references aren't so lucky.  MWE:  \documentclass[letterpaper, 12pt, twoside]{exam}  \usepackage{hyperref}  \newcounter{KSRLessonNumberCounter} \setcounter{KSRLessonNumberCounter}{1} \newcommand{\KSRSection}[2][\value{KSRLessonNumberCounter}]{         \setcounter{KSRLessonNumberCounter}{#1}         \section*{Lesson \arabic{KSRLessonNumberCounter}: #2}         \addcontentsline{toc}{section}{Lesson \arabic{KSRLessonNumberCounter}: {#2}}         \stepcounter{KSRLessonNumberCounter} } \newcommand{\KSRSubsection}[1]{         \subsection*{#1}         \addcontentsline{toc}{subsection}{#1} } \newcommand{\KSRLessonNumber}{ }  \begin{document}  \KSRSection{Sample}  This is the first lesson. The second lesson is \nameref{sec:blah}.  \KSRSection{Blah} \label{sec:blah}  This is the second section  \KSRSection{Lorem Ipsum}  Dollars are Set in \nameref{sec:blah}.  \end{document} Sequence of differentiable functions Let $f_n$ be a sequence of differentaible functions on $[0,1]$ to $\mathbb{R}$ converging uniformly to a function $f$ on $[0,1]$, Then    $f$ is differentiable and Riemann integrable there $f$ is uniformly continuous and R-integrable $f$ is continuous, need not be differentiable on $(0,1)$ and need not be R-integrable on $ [0,1]$, $f$ need not be continuous.   Well I think 2 is only correct statement  as every $f_n$ are differentiable on $[0,1]$ hence uniformly contnous, and as they converge uniformly to $f$ on $[0,1]$ so $f$ is uniformly continuous and every continuous function is rieman integrable..Am I correct? I'm trying to prove that the converse of Lagrange's theorem is true for finite abelian groups (i.e. "given an abelian group $G$ of order $m$, for all positive divisors $n$ of $m$, $G$ has a subgroup of order $n$"). This is an exercise from a book, and it is in the section on finite abelian groups, so I know I have to use the fundamental theorem of finite abelian groups. I have come up with a proof, but it seems a bit messy, and I'm not entirely sure if it's correct. It is given below.     Let the order of $G$ be $m$ = $p_1^{\alpha_1} \ldots p_k^{\alpha_k}$. It is known that $G$ is a direct product of $p$-groups, say:      $$G = G_1 \times \ldots \times G_k$$      where each $G_i$ is a $p_i$-group. By the fundamental theorem of finite abelian groups, each $G_i$ is isomorphic to a direct product of cyclic groups of the form      $$\mathbb{Z}_{{p_i}^{\beta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\beta_l}},$$      where $\beta_1, \ldots, \beta_l$ are positive integers such that $\sum_{j=1}^l \beta_j = \alpha_i$. Now if $n$ divides $m$, then we must have      $$n = p_1^{\gamma_1} \ldots p_k^{\gamma_k}$$      for some $\gamma_1, \ldots, \gamma_k$ with $0 \leq \gamma_i \leq \alpha_i$.      Claim: Each $G_i$ has a subgroup of order $p_i^{\gamma_i}$      Proof: As above, we have that       $$ G_i \cong \mathbb{Z}_{{p_i}^{\beta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\beta_l}} $$      where $\beta_1, \ldots, \beta_l$ are positive integers such that $\sum_{j=1}^l \beta_j = \alpha_i$.      Now since $0 \leq \gamma_i \leq \alpha_i$, we can find   $l$ numbers $\delta_1, \ldots , \delta_l$ such that $\gamma_i = \sum_{j=1}^l \delta_j$, and $0 \leq \delta_j \leq \beta_j$. (This choice of numbers is not necessarily unique).      Then $p_i^{\delta_j} | p_i^{\beta_j}$ for each $j = 1, \ldots , l$. Hence, for each factor $\mathbb{Z}_{{p_i}^{\beta_j}}$, there exists a subgroup of order $p_i^{\delta_j}$, namely $\mathbb{Z}_{{p_i}^{\delta_j}}$ (using the fact that the converse of Lagrange's theorem is true for finite cyclic groups). Taking the direct product of each of these subgroups, we get a new subgroup $G_i'$ of $G_i$:      $$G_i' \cong \mathbb{Z}_{{p_i}^{\delta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\delta_l}}$$      The order of this subgroup is $p_i^{\delta_1} \times \ldots \times p_i^{\delta_l} = p_i^{\delta_1 + \ldots + \delta_l} = p_i^{\gamma_i} $. So we have found a subgroup of $G_i$ of order $p_i^{\gamma_i}$, as required.      So each factor $G_i$ in the product $G = G_1 \times \ldots \times G_k$ has a subgroup $G_i'$ of order $p_i^{\gamma_i}$.       Therefore, $G$ has a subgroup   $$G_1' \times G_2' \times \ldots \times G_k'$$       of order $p_1^{\gamma_i}...p_k^{\gamma_k} = n$, which completes the proof.   I have two questions about this: firstly, does this proof seem to work? Secondly, is there a way to make the proof more concise (e.g. a way to prove the statement without using all these indices)? Increase iPhone call history Can you 'exit' a loop in PHP? Create SOAP envelope of XML and send it as HttpWebRequest to WebService Are mounts proficient in armour (barding)? How to find the 'sizeof' (a pointer pointing to an array)? First off, here is some code:  int main()  {     int days[] = {1,2,3,4,5};     int *ptr = days;     printf("%u\n", sizeof(days));     printf("%u\n", sizeof(ptr));      return 0; }   Is there a way to find out the size of the array that ptr is pointing to (instead of just giving its size, which is four bytes on a 32-bit system)? If A is invertible, prove that $\lambda \neq 0$, and $\vec{v}$ is also an eigenvector for $A^{-1}$, what is the corresponding eigenvalue? If A is invertible, prove that $\lambda \neq 0$, and $\vec{v}$ is also an eigenvector for $A^{-1}$, what is the corresponding eigenvalue?  I don't really know where to start with this one. I know that $p(0)=det(0*I_{n}-A)=det(-A)=(-1)^{n}*det(A)$, thus if both $p(0)$ and $det(0) = 0$ then $0$ is an eigenvalue of A and A is not invertible. If neither are $0$, then $0$ is not an eigenvalue of A and thus A is invertible. I'm unsure of how to use this information to prove $\vec{v}$ is also an eigenvector for $A^{-1}$ and how to find a corresponding eigenvalue. Will the Windows 10 Free Upgrade stay in the Professional channel? Cannot change lock screen background on Ubuntu 18.04 Is there a single word to describe a solution that hasn't been optimized? I am working with some code, and I would like to describe the difference in its performance when it is optimized versus when it is not optimized. Unfortunately, I can't find a word to describe the code before it is optimized; "sub-optimal" might work, but I want to emphasize that this is the first iteration of the code. If "unoptimized" were a word, I would use that, but I haven't found any references to it in any dictionary.  Perhaps "raw" or "original"? How do I reset my manor so I can rebuild it with different options?