How to remove author name default capitalization? According to biblatex doc (p.80-81) :     \Parencite is similar to \parencite but capitalizes the name prefix of   the first name in the citation if the useprefix option is enabled,   provided that there is a name prefix and the citation style prints any   name at all.      \Textcite is similar to \textcite but capitalizes the name prefix of   the first name in the citation if the useprefix option is enabled,   provided that there is a name prefix.   My problem is that \parencite and \textcite capitalize the author name by default. Therefore, there is no difference between these commands (\textcite has the same behavior than \Textcite). I am using authoryear-comp style.  It it a big problem for me because I cannot cite using \textcite. Indeed it will produces something like "according to NAME (2008), there is ...". The capitalized author name is ugly.  MWE :   \RequirePackage{filecontents}  \begin{filecontents}{\jobname.bib} @Book{author00:_title,   author =   {Author},   title =    {Title},   publisher =    {Publisher},   year =     2000} \end{filecontents}  \documentclass[12pt,twoside,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[english,french]{babel}  \usepackage{lmodern} \usepackage{xspace}  \usepackage[autostyle=true]{csquotes} \usepackage[style=authoryear-comp,      hyperref,      backend=biber,      isbn=false,      doi=false,      url=false,      ]{biblatex} \AtEveryBibitem{\clearlist{language}} \renewcommand*{\newunitpunct}{\addcomma\space} \addbibresource{\jobname.bib} \usepackage[colorlinks]{hyperref}  \begin{document}  \noindent \verb+\parencite{author00:_title}+ :  \parencite{author00:_title}\\ \verb+\Parencite{author00:_title}+ : \Parencite{author00:_title}\\ \verb+\textcite{author00:_title}+ : \textcite{author00:_title}\\ \verb+\Textcite{author00:_title}+ : \Textcite{author00:_title}  %\printbibliography \end{document}
Changing default \citet font in biblatex I'm using biblatex to manage my citation in a natbib style :  \usepackage[style=alphabetic,maxnames=6,natbib=true]{biblatex}   However when I use the \citet command, the author's name is in a \textsc font, I would like to change this behavior so that the font used is the default one, but I could not find how to do this in biblatex's documentation.
How can I dual boot Windows and Ubuntu? I am foreign graduate student in USA. I recently completed my undergrad (from USA) and have a bit of money saved up. I know that F1 students cannot work off campus and other business activities such stock trading etc are prohibited as well. I wanted to know if opening an account in an investment management firm like Charles Schwabb or Betterment is allowed. I would not be working but I will still be earning a small interest on my deposits. If that is not allowed, are there any good and legal ways for foreign students to earn a little interest on savings? Why is 'that' sometimes optional before dependent clauses? What is the "show desktop" keyboard shortcut? I just upgraded to 11.10 from 11.04. Previously the shortcut to show the desktop was Super +D (same as in Windows), But now it is not working. What is the new shortcut? How to repair a corrupted Excel file I'm trying to restore some files for a friend who messed up a USB stick by pulling it out of the computer too soon. By searching invisible partitions I was able to restore the files. All files are intact, except 1 excel file that seems to be corrupted.  I've googled about ways to repair an Excel file, because the default repair options in Excel 2003-2010 do not work for this file. I've came across a numerous  third-party apps, of which the  one seemes to be able to do the job: Recovery Toolbox for Excel.  Of-course, this and all other apps I found require a purchase to actually repair the file. It seems a bit silly to pay 30 dollars for a one time thing, and something that is not for myself. So, I would like to keep that as a last-resort option.  Is there anything I can try to fix this? I've attached a screenshot of how the file looks when I open it on my OSX when converted to a HTML file (to view the code), for  the option that it could be a character set problem. Prove that if $n$ is a composite, then $2^n-1$ is composite. Not sure if I'm doing this correctly but this is what I've done:  Assume that $n$ is composite and suppose $2^n-1$ is a prime for $n \gt 2$. Then, $2^n-1 = 2k$ for some $k \in \Bbb Z $, $\forall n$.  But this would be a contradiction since $n \gt 2$.  I'm not sure if this a correct proof.  EDIT: I realized that it's to prove $2^n - 1$ NOT $2^{n-1}$ so my attempt above is completely wrong. Overloading getter and setter causes a stack overflow in C# Are we slightly lighter during the day and slightly heavier at night, owing to the force of the Sun's gravity? How unique is the php session id How unique is the php session id? I got the impression from various things that I've read that I should not rely on two users never getting the same sessionid. Isn't it a GUID? I work in D8, I create a custom module first_module. By modules create a url and display all nodes by *.html.twig file.  Write code in controller.php  public function contentthree() {    $connection = Database::getConnection();    $query = $connection-&gt;select('node', 'n');   $query-&gt;fields('n', ['nid', 'type']);    $groups = $query-&gt;execute()-&gt;fetchAll();    foreach ($groups as $group) {     $events[] = [$group-&gt;nid, $group-&gt;type];   }    return [     '#theme'  =&gt; 'first_module_display',     '#events' =&gt; $events,   ]; }   In *.module  function first_module_theme($existing, $type, $theme, $path) {    return [     'first_module_display' =&gt; [       'variables' =&gt; ['events' =&gt; NULL],       'template'  =&gt; 'first-module-list',     ],   ]; }  function _get_node_details() {    $output = 'Welcome D8';    $output .= 'Node Details';    return $output; }   and create templates first-module-list.html.twig and display data  {%for key, data in events %}  {%for key1, data1 in data %}     {{ data1 }}  {% endfor %}  {% endfor %}   work fine and showing data correct. But, my question is how to call function _get_node_details() and how to render node details in first-module-list.html.twig files? Do Dragonborn get tails/wings? 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! Compactness of $(M, d)$ in terms of continuous functions $f:M\to \mathbb R$ Suppose $(M,d)$ a metric space. I want to show that      If every continuous real-valued function on $M$ attains a maximum, then $(M, d)$ is compact.   I was trying to do this by assuming $M$ non-compact and showing explicitly a function which does not attain a maximum. If $M$ is not bounded it's enough to take the distance from a fixed point as we can always find a sequence which "exits" every compact and pushes the distance at infinity. But what if $M$ is bounded? Debugging errors of SAGA interpolation algorithms in QGIS Processing? Are universal power supplies for laptops worth it? My laptop's cable broke and sadly it seems like I need to buy a new power supply.  What I wonder is, I've found that universal power supplies are much cheaper than the official HP ones. Are they really worth it?  Another question: the original supply has these specs: 65W, 18.5V, 3.5A. However the universal supplies are more powerful (90W, 100W) and come in integer voltages (17V, 18V, 19V, etc). Are they compatible? Should I set them to 18V or 19V? I understand that the eigenvectors of a linear transformation are the vectors that, when transformed, are simply scaled by a factor of the corresponding eigenvalue. Therefore $$ A\vec{x}=\lambda\vec{x}. $$ We can replace the scaling of the vector $\vec{x}$ by $\lambda$ with its equivalent linear transformation, $\lambda I$. Then we can rearrange this equation as follows: $$ A\vec{x}-(\lambda I)\vec{x}=\begin{bmatrix}0\\0\end{bmatrix}\\  (A-\lambda I)\vec{x}=\begin{bmatrix}0\\0\end{bmatrix}\\  $$ For this to be true either $\vec{x}$ must be a zero vector or the transformation $A-\lambda I$ must have $\vec{x}$ in its null-space. The course I'm following says that we can solve for $\lambda$ by finding when the determinant of $A-\lambda I$ is zero, but why does that guarantee that $\vec{x}$ will become the zero vector when transformed by $A-\lambda I$? I understand that for $\vec{x}$ to be an eigenvector this transformation must have a determinant of zero, but I don't see how the inverse can be true; having a determinant of zero doesn't gaurantee that $\vec{x}$ will be transformed into the zero vector. For example, let $$ (A-\lambda I) = \begin{bmatrix}1&amp;2\\0&amp;0\end{bmatrix}, \vec{x}=\begin{bmatrix}1\\1\end{bmatrix} $$ Then, although $\det\begin{bmatrix}1&amp;2\\0&amp;0\end{bmatrix} = 0$, it does not gaurantee $(A-\lambda I)\vec{x}$ to be the zero vector. $$ \begin{bmatrix}1&amp;2\\0&amp;0\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\0\end{bmatrix}  $$ We can see that the transformation gives us $\begin{bmatrix}3\\0\end{bmatrix}$ and not the zero vector. Why does finding when the determinant of this transformation is equal to zero help us find the eigenvectors? Am I missing something? I am using nginx/0.7.68, running on CentOS, with the following configuration:  server {     listen       80;     server_name ***;     index index.html index.htm index.php default.html default.htm default.php;      location / {             root   /***;             proxy_pass   http://***:8888;             index  index.html index.htm;     }     # where *** is my variables   The proxy_pass is to a DNS record whose IP changes frequently. Nginx caches the outdated IP address, resulting in a request to the wrong IP address.  How can I stop nginx from caching the IP address, when it is outdated? What is the purpose of the EBP frame pointer register? Is it true every ideal of $\mathbb{Z}[x]$ is finitely generated? Let $I$ be any ideal of $\mathbb {Z}[x]$, and $I=\left&lt;{f_1,f_2 , ...}\right&gt;$.  Without loss of generality, $f_1$ has minimal (absolute value) coefficient of constant term (if there exist).  Then, let $g$ be any element of $I$ .  We have for some integer $m$ that $g - mf_1$ is element of $I$ (minimal coefficient and division algorithm).  I guess this process terminate finitely, but I don't know how can I do next step.  ( ex. $g-mf_1, g-mf_1-nf_2, ...$)  Please help me and thanks for in advance. Are enemies with immunity to the Poisoned condition unaffected by Contagion? After the , the spell  now inflicts the  condition upon a successful melee spell attack. The relevant spell text is below, with updated portions highlighted in bold:  Your touch inflicts disease. Make a melee spell attack against a creature within your reach. On a hit, the target is poisoned. At the end of each of the poisoned target's turns, the target must make a Constitution saving throw. If the target succeeds on three of these saves, it is no longer poisoned, and the spell ends. If the target fails three of these saves, the target is no longer poisoned, but choose one of the diseases below. The target is subjected to the chosen disease for the spell's duration. Since this spell induces a natural disease in its target, any effect that removes a disease or otherwise ameliorates a disease's effects apply to it.  The wording of the spell leads me to conclude that this spell still inflicts a disease, which has the immediate effect of poisoning the target. Therefore, even if a target has poison immunity, they will still make Constitution saves as designated to see if the disease takes effect. However, the line, &quot;At the end of each of the poisoned target's turns...&quot; also leads me to think that an enemy with poison-immunity cannot be the Poisoned target, therefore disallowing the remaining sequence of events. Are targets with immunity to the Poisoned condition therefore immune to contagion?