I am trying to make the title page for my master thesis. I found on the internet a template, in which the cover page is given by the following code  \begin{titlepage}  \begin{center}  \textsc{\LARGE University Name}\\[1.5cm]  \textsc{\Large Master Thesis}\\[0.5cm]    {\huge \bfseries Thesis Title}\\[0.4cm]    \begin{minipage}{0.4\textwidth}  \begin{flushleft} \large  \emph{Author: Name}  \end{flushleft}  \end{minipage}   \begin{minipage}{0.4\textwidth}  \begin{flushright} \large  \emph{Supervisor: Name} \end{flushright}  \end{minipage}\\[3cm]     \large \textit{Department of Economics}\\[0.4cm]  \textit{Uppsala}\\[0.4cm]   {\large \today}\\[4cm] % Date %\includegraphics{Logo}   \vfill  \end{center} \end{titlepage}   When I compile latex in Lyx, my name and my supervisor's name are aligned in the center and not in the left and right, as they supposed to be  What I am missing?  PS: Here is the link
I'm trying to get my csv table aligned to the left of my graph. The graph is appearing below (and indented) the table. Any ideas?  \begin{minipage}{0.5\textwidth} \pgfplotstabletypeset[col sep=comma, columns/distance/.style={column name={Col1}, column type={|c|}}, columns/loss/.style={column name = {Col2},column type={c|}}, every head row/.style={before row=\hline}, after row={\hline} ]{graph-data/data.csv} \end{minipage}  \begin{minipage}{0.5\textwidth} \begin{tikzpicture} \begin{axis}[ xlabel = Distance ($m$), ylabel = Loss ($dB$)] \addplot table [col sep=comma]{graph-data/data.csv}; \end{axis} \end{tikzpicture} \end{minipage}
Why does changing the returned variable in a finally block not change the return value? Can't install apps from App Store on iPhone with iOS 7 apache2 mod_rewrite redirect without http body I am trying to find the name of a direct to TV movie, possibly a 2-3 part miniseries which I believe aired on the SciFi channel in the 90s. I don't remember who starred in it, but the movie opened with an astronaut attempting reentry whose shuttle ends up exploding. He awakes in a large bubble of air in a river and after reaching the surface and swimming to shore, meets several other people, from seemingly many different times throughout history (including one alien, who died on Earth).  Over the course of the movie, we learn the world is the afterlife, where everyone goes when they die. The major plot involves launching a steamboat (built by Mark Twain, no less) to travel up a river to the mountains, where the beings responsible for bringing everyone to this world are.  Does anyone know what the title of this movie/miniseries is? I'm interested in Gaussian blur layer that blurring all inside of some figure located on this layer.  I mean, blurred background of some exact area. And in case of I move this area, blurred image must be updated. If I move some object in background - blur must be updated. etc. OpenCV argc and argv confusion In the new canon what do we actually know about Darth Plagueis? I read and loved  but with all the recent changes to what is and isn't canon I'm losing track.  There's a common theory that Supreme Leader Snoke is Darth Plagueis, most are discounted because Plagueis is a Muun. But where does that information come from? What current canon information do we have about Plagueis that disproves this theory and what has been marked as Legends? I read many places that the rule of thumb for swap space is to double the amount of physical RAM. However, 32 GB does seem a LOT. Do I need that much? Do I need it at all with this high amount of physical RAM? Show a connected metric space (X,d) is $\epsilon$-chainable for $\epsilon &gt;0$       $\epsilon$-chainable Definition      (X,d) is $\epsilon$-chainable if given any two points $a,b \in X$, $\exists$ a finite set $\{ x_0,x_1,...,x_n\}$ of points in X with $x_0 = a$, $x_n=b$ and $d(x_{n-1},x_n)&lt;\epsilon$, $\forall$ n      Clopen Definition      A set is clopen $iff$ its boundary is empty      Connected Metric Definition      Let $(X,d)$ be a metric space. It is connected $iff$ it has no proper subsets which are clopen      The following are equivalent:         $X$ is connected   The only subsets of $X$ that are clopen are $\emptyset$ and $X$   $\nexists$ non-empty, disjoint, open sets $A$ and $B$ such that $X=A\bigcup B$   $\nexists$ non-empty, disjoint, closed sets $A$ and $B$ such that $X=A\bigcup B$         Theorem      Let $(X,d)$ be a metric space, and let $S$ be any subset of $X$. Then $S$ is closed $iff$ $S^c$ is open.     So how would someone show this exactly?  Let the set $A$ be the set of all elements in $(X,d)$ that can be reached from $a\in X$ in a finite number of steps with length less than $\epsilon$.  Does this mean that there should be no isolated points? (an isolated point would be at distance infinity from any other point, or distance zero from itself, since $\epsilon &gt;0$ we don't want this?)  Knowing that $X$ is connected $\implies$the only subsets of X that are clopen are $\emptyset$ and $X$  Also, by construction of A;   A is the set of all $\epsilon$-chains between any element and $a$  And, so $A^c = \emptyset$ if $A$ was $\epsilon$-chainable (because A has to have a path from all elements to a, for this to be true, the complement wouldn't have anything in it)  Leaving only $A$ as the remaining clopen set  Does this make any sense? Or should an alternative be used, and how to formalize it.  Thanks. Fix a wobbly Shimano freehub? Find number of files with a specific extension, in all subdirectories Is there a way to find the number of files of a specific type without having to loop through all results inn a Directory.GetFiles() or similar method? I am looking for something like this:  int ComponentCount = MagicFindFileCount(@"c:\windows\system32", "*.dll");   I know that I can make a recursive function to call Directory.GetFiles , but it would be much cleaner if I could do this without all the iterating.  EDIT: If it is not possible to do this without recursing and iterating yourself, what would be the best way to do it? I'm using a class I've derived from DocumentPaginator (see below) to print simple (text only) reports from a WPF application. I've got it so that everything prints correctly, But how do I get it to do a print preview before printing? I have a feeling I need to use a DocumentViewer but I can't figure out how.  Here's my Paginator Class:  public class RowPaginator : DocumentPaginator {     private int rows;     private Size pageSize;     private int rowsPerPage;      public RowPaginator(int rows)     {         this.rows = rows;     }      public override DocumentPage GetPage(int pageNumber)     {         int currentRow = rowsPerPage * pageNumber;         int rowsToPrint = Math.Min(rowsPerPage, rows - (rowsPerPage * pageNumber - 1));         var page = new PageElementRenderer(pageNumber + 1, PageCount, currentRow, rowsToPrint)                        {                            Width = PageSize.Width,                            Height = PageSize.Height                        };         page.Measure(PageSize);         page.Arrange(new Rect(new Point(0, 0), PageSize));         return new DocumentPage(page);     }      public override bool IsPageCountValid { get { return true; } }      public override int PageCount { get { return (int)Math.Ceiling(this.rows / (double)this.rowsPerPage); } }      public override Size PageSize     {         get { return this.pageSize; }         set         {             this.pageSize = value;             this.rowsPerPage = PageElementRenderer.RowsPerPage(this.pageSize.Height);             if (rowsPerPage &lt;= 0)                 throw new InvalidOperationException("Page can't fit any rows!");         }     }      public override IDocumentPaginatorSource Source { get { return null; } } }   The PageElementRenderer is just a simple UserControl that displays the data (at the moment just a list of rows).  Here's how I use my Row Paginator  PrintDialog dialog = new PrintDialog(); if (dialog.ShowDialog() == true) {     var paginator = new RowPaginator(rowsToPrint) { PageSize = new Size(dialog.PrintableAreaWidth, dialog.PrintableAreaHeight) };      dialog.PrintDocument(paginator, "Rows Document"); }   Sorry for the code dump, but I didn't want to miss something relevant. does anyone know the following result? If it holds in this form and any source which presents it? Thanks a lot.  Consider metric space $(X,d_{X})$.  Let $f:A \subset (X,d_{X}) \rightarrow \mathbb{R}$ be a Lipschitz continuous function i.e. $\exists K &gt; 0$ such that  \begin{align*} |f(x)-f(y)| \leq Kd_{X}(x,y) ~~~~\forall x,y \in A \end{align*}  then define functions  $$\bar{f_{1}}(x) := \sup\lbrace f(y) - Kd_{X}(x,y): y\in A \rbrace ~~ \forall x  \in X$$ and $$\bar{f_{2}}(x) := \inf\lbrace f(y) + Kd_{X}(x,y): y\in A \rbrace ~~ \forall x  \in X$$ it follows that  \begin{align*} \bar{f_{1}}, \bar{f_{2}}: X \rightarrow \mathbb{R}  \end{align*} are Lipschitz continuous with  \begin{align*} |\bar{f_{i}}(x)-\bar{f_{i}}(z)| \leq Kd_{X}(x,z) ~~~\forall x,z \in X \text{ and } \forall  i = 1,2 \end{align*} and  \begin{align*} \bar{f_{i}}(x) = f(x) ~~~ \forall x \in A \text{ and } \forall i =1,2 \end{align*} The internals of bcrypt I'm attempting to get a better understanding of cryptography and have run into some questions about bcrypt.   Bcrypt relies on the blowfish cipher, which is a reversible method of encryption. But bcrypt is not reversible. How is that possible? Or am I mistaken? What does bcrypt do on top of blowfish to make it irreversible? Why does bcrypt use blowfish instead of twofish or threefish? After a company has issued shares, why should it care if the value of the shares fluctuate? Since the shares are now publicly traded, if the value fluctuates, that is strictly a transaction between buyer and seller, none of whom are directly related to the company. E.g. if I sell X an iPhone for $100, why should I care if X is selling the iPhone to Y for 1 dollar or 1000 dollars? I have got my $100, so I need not bother. No? A linear operator commuting with all such operators is a scalar multiple of the identity. The question is from Axler's "Linear Algebra Done Right", which I'm using for self-study.  We are given a linear operator $T$ over a finite dimensional vector space $V$. We have to show that $T$ is a scalar multiple of the identity iff $\forall S \in {\cal L}(V), TS = ST$. Here, ${\cal L}(V)$ denotes the set of all linear operators over $V$.  One direction is easy to prove. If $T$ is a scalar multiple of the identity, then there exists a scalar $a$ such that $Tv = av$, $\forall v \in V$. Hence, given an arbitrary vector $w$, $$TS(w) = T(Sw) = a(Sw) = S(aw) = S(Tw) = ST(w)$$ where the third equality is possible because $S$ is a linear operator. Then, it follows that $TS = ST$, as required.  I am, however, at a loss as to how to tackle the other direction. I thought that a proof by contradiction,  ultimately constructing a linear operator $S$ for which $TS \neq ST$, might be the way to go, but haven't made much progress.   Thanks in advance! Why did Douglas Adams pick 42 as the ultimate answer? Star Trek often refers to 47 and Star Trek made a sport out of finding real or imagined references to 47. Why did Douglas Adams pick 42 as the ultimate answer in the Hitchhikers guide to the galaxy? Official reasons for "Software caused connection abort: socket write error" I have 10 years of daily returns data for 28 different currencies. I wish to extract the first principal component, but rather than operate PCA on the whole 10 years, I want to rollapply a 2 year window, because the currencies' behaviours evolve and so I wish to reflect this. However I have a major problem, that is that both the princomp() and prcomp() functions will often jump from positive to negative loadings in adjacent PCA analyses (ie 1 day apart). Have a look at the loading chart for the EUR currency:    Clearly I can't use this because adjacent loadings will jump from positive to negative, so my series which uses them will be erroneous. Now take a look at the absolute value of the EUR currency loading:    The problem is of course that I still cannot use this because you can see from the top chart that the loading does go from negative to positive and back at times, a characteristic which I need to preserve.   Is there any way I can get around this problem? Can I force the eigenvector orientation to always be the same in adjacent PCAs?   By the way this problem also occurs with the FactoMineR PCA() function. The code for the rollapply is here:  rollapply(retmat, windowl, function(x) summary(princomp(x))$loadings[, 1], by.column = FALSE, align = "right") -&gt; princomproll Divisibility is transitive: $\ a\mid b\mid c\,\Rightarrow\ a\mid c$ As the title says, if a number is divisible by a number, is it always divisible by that number's factors?  An example being that $100$ is divisible by $20$, it is also divisible by $10, 5, 4, 2$ as well?  Does this always apply?