Is this that simple?
How do I install Ubuntu?
I want to use the top half of the circle to make a cut but whenever I apply the boolean modifier, the circle just stays there and becomes attached to my other object.  I've managed to make the cut in a standard cube but not sure why it's not working for this...? Splitting string into pair of characters in Ruby Ubuntu 11.10 is installed, and I have updated my system via Ubuntu Software Centre, and after the update 3 updates remain in the list. But I am unable to select them, the tickboxes cannot be checked and Ubuntu Software Centre gives no error when I try to check.  Also, below it says "There is no update to be installed" and even though there seems to be updates Install Updates button is unclickable.  What should I do in order to install these updates? In Greek, λογικοφανής is a word that describes a person who spits out seemingly logical statements. It often has a negative connotation as it is targeted towards people who tend to fool the public with supposedly beautifully-constructed thoughts that sound convincing in our ears but are not necessarily accurate. The first word that comes to mind is deceptive, but it is not quite what I look for. Any thoughts? How do per-site Metas differ from Meta Stack Exchange, and where should I go to ask my question? How do per-site metas, such as Meta Writers Stack Exchange or Meta Stack Overflow differ from the general Meta Stack Exchange?  Where should I go to ask my question? Problem 4.3.32 in Linear Algebra, Lay:  Let $V$ and $W$ be vector spaces, let $T:V\to W$ be a linear transformation, and let $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ be a subset of $V$. Suppose that $T$ is a one-to-one transformation [...]. Show that if the set of images $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set.  All I can think of doing: If $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then there exist weights $c_1, \dots, c_p$, not all zero, so that $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = \mathbf{0}$$ Since $T$ is linear: $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = T(c_1\mathbf{v}_1) + \cdots + T(c_p \mathbf{v}_p)=T(c_1 \mathbf{v}_1 +\cdots + c_p \mathbf{v}_p)=\mathbf{0}$$ So the set $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ must be linearly dependent. Am I missing something? And why must $T$ be on-to-one for this to be true? An example showing that it is not true if $T$ is not one-to-one would be great! How to manage a redirect request after a jQuery Ajax call I'm using $.post() to call a servlet using Ajax and then using the resulting HTML fragment to replace a div element in the user's current page. However, if the session times out, the server sends a redirect directive to send the user to the login page. In this case, jQuery is replacing the div element with the contents of the login page, forcing the user's eyes to witness a rare scene indeed.   How can I manage a redirect directive from an Ajax call with jQuery 1.2.6? Nautilus does not show the left bar, how do I turn it back on? How to get Drupal 7 installed in a sub-folder / sub-directory properly? Currently i'm trying to installing a drupal 7 installation in a subfolder of another drupal public_html folder, but it doesnt seem to work. I am getting a redirect loop error in chrome. I already modified the .htaccess in the parent drupal installation to ignore the sub-folder and i also set the $base_url in the sub-directory drupal installation. QGIS creating a fixed distance buffer over selected areas not the entire layer what im doing is making a fixed distance buffer and it's working splendidly but i need the buffer to be created over certain selected areas not the entire layer, and at the creation wizard is only offers the the option of selecting the entire layer, and i need some areas within that layer to have a different buffer and some none at all.  is there a way around that? alias to take a parameter and exit after execution Help center still has the guidance about old Favorite tags Shape of confidence interval for predicted values in linear regression I have noticed that the confidence interval for predicted values in an linear regression tends to be narrow around the mean of the predictor and fat around the minimum and maximum values of the predictor. This can be seen in plots of these 4 linear regressions:    I initially thought this was because most values of the predictors were concentrated around the mean of the predictor. However, I then noticed that the narrow middle of the confidence interval would occur even if many values of were concentrated around the extremes of the predictor, as in the bottom left linear regression, which lots of values of the predictor are concentrated around the minimum of the predictor.  is anyone able to explain why confidence intervals for the predicted values in an linear regression tend to be narrow in the middle and fat at the extremes? Hartshorne's Algebraic Geometry defines an $\mathcal O_X$-module $\mathscr F$ to be quasi-coherent if there is an open affine cover $(U_i=\operatorname{Spec} A_i)_{i\in\mathcal I}$ of $X$ such that each $\mathscr F\rvert_{U_i}$ is isomorphic to the sheaf of modules $\widetilde{M_i}$ associated to some $A_i$-module $M_i$. It is later proved that this is equivalent to the statement that for any open affine $U=\operatorname{Spec} A\subseteq X$, there exists an $A$-module $M$ with $\mathscr F\rvert_U\cong\widetilde M$.  On the other hand, Grothendieck's Éléments de Géométrie Algébrique defines $\mathscr F$ to be quasi-coherent if every point $x\in X$ admits an open neighborhood $U\subseteq X$ such that $\mathscr F\rvert_U$ is isomorphic to the cokernel of a morphism of free $\mathcal O_U$-modules.  Now I wonder how those definitions are equal. I tried proving it by myself, but both directions seem quite nontrivial to me... Any hints? Why do the planets' orbital distances fall on an exponential curve? Background: I was recently reading a book on the planets to my son and I noticed a pattern in the distributions of the planets.  The planets' distances roughly follow an exponential distribution.  Below is a plot of scaled, log orbital distances $$ \tilde{d_n} = \frac{\log(d_n/d_1)}{\log(d_2/d_1)} $$ with the line $an+b$:   Where $d_1$ corresponds to Mercury and so on. Ceres is included, Pluto is excluded.  By linear regression, $a = 0.90, b = -1.06$.  For the statistically minded, the data has a Pearson's correlation of 0.996.  Note that this is a well known phenomenon, see  and references.  The code used to generate the plot may be provided on request.  Question: What is the mechanism that leads to this distribution?  Aside: Is there a good introductory text on solar system formation for the mathematically inclined?  Update: This is also known as Titius–Bode law. Do the Ents go to the Undying Lands? Tufte advocates to "minimize the ink-to-data ratio" for charts. I am trying to style a multi data series bar chart, that can have a grid and axes as shown below (chart at the left is from the Economist and on the right my attempt).      Here is the code:  \documentclass[justified]{tufte-book} \usepackage{pgfplots,lipsum} \pgfplotsset{compat=newest} \begin{document} \section{Bar Charts}  \lipsum[1-2]\vfill   \begin{figure*}[htbp]   \fbox{  \begin{tikzpicture}   \centering   \begin{axis}[         ybar,           title={Cumulative Progress of Works},         height=8cm,         width=15.5cm,         bar width=0.4cm,         ymajorgrids, tick align=inside,         enlarge  y limits={value=.2,auto}, % see the upper          ymin=0,         ymax=100,         axis x line=bottom,          axis y line=right,         %enlarge x limits =e,         legend style={at={(0.5,-0.2)},         anchor=north,legend columns=-1},         ylabel={Percentage (\%)},         symbolic x coords={            Sep-11,Oct-11,Nov-11,Dec-11,            Jan-12,Feb-12,            Mar-12,           Apr-12},        xtick=data,        nodes near coords,        every node near coord/.append style={         anchor=mid west,         rotate=70       }     ]     \addplot coordinates {       (Sep-11,75.4064)       (Oct-11, 72.7961)        (Nov-11,94.4597)       (Dec-11,66.6786)        (Jan-12,67.5600)        (Feb-12,88.2339)       (Mar-12,78.6138)        (Apr-12,58.9129) };    \addplot coordinates {       (Sep-11,75.4064)       (Oct-11, 89.7961)        (Nov-11,94.4597)       (Dec-11,76.6786)        (Jan-12,77.5600)        (Feb-12,78.2339)       (Mar-12,88.6138)        (Apr-12,78.9129) };    \addplot coordinates {       (Sep-11,75.4064)       (Oct-11, 89.7961)        (Nov-11,94.4597)       (Dec-11,76.6786)        (Jan-12,77.5600)        (Feb-12,78.2339)       (Mar-12,88.6138)        (Apr-12,78.9129) };      \legend{First Fix,Second Fix,Third Fix}   \end{axis}   \end{tikzpicture} }  \caption{\protect\raggedright Cumulative progress for all MEP works. Notice the slower rate of production during the last three months.} \end{figure*}  \lipsum[1-3] \end{document}   I have moved the y-axis to the right, but cannot get rid of the axis line. The plot is also not fitting within the plot width, despite using \enlarge.   Edit  And thanks to Jake the ugly duckling has been transformed to the following, with minor variations to the code as suggested by Jake. I have only added a stronger color for one of the bars in order to highlight it: The following code helps me validate an XML file with an XSD schema.  XmlReaderSettings settings = new XmlReaderSettings(); settings.Schemas.Add(null, xsdFilePath); settings.ValidationType = ValidationType.Schema; settings.ValidationEventHandler += new System.Xml.Schema.ValidationEventHandler(settings_ValidationEventHandler); XmlDocument document = new XmlDocument(); document.Load(xmlFilePath); XmlReader rdr = XmlReader.Create(new StringReader(document.InnerXml), settings);  while (rdr.Read()) {  } isValid = true;   The ValidationEventHandler also tells me what the errors are, but doesn't tell me on 'where' or 'on which line' they are located. Is there any way to get the line number where the XML fails to be validated? $X_1, X_2, \ldots, X_n$ are $n$ i.i.d. uniform random variables. Let $Y = \min(X_1, X_2,\ldots, X_n)$. Then, what's the expectation of $Y$(i.e., $E(Y)$)?  I have conducted some simulations by Matlab, and the results show that $E(Y)$ may equal to $\frac{1}{n+1}$. Can anyone give a rigorous proof or some hints? Thanks! diff a ruby string or array How do I do a diff of two strings or arrays in Ruby?