splash screen comes up then goes to a blank screen after updating software. How do I access my desktop?? Desperate for help! Thanks
I am trying to boot Ubuntu on my computer.  When I boot Ubuntu, it boots to a black screen.  How can I fix this? Table of Contents:
Fast calculation for $\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0$ I want to show that $\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0$, but is there a faster method than finding the contour and doing all computations?  Otherwise my idea is to do the substitution $x=e^t$, integral than changes to $\int _{-\infty}^{\infty}\frac{t e^t}{1+e^{2t}}dt$. Next step is to take the contour $-r,r,r+i\pi,-r+i\pi$ and integrate over it... Consider the  function     $$ r(x) = \mathbb{E}(Y \mid X = x) $$   This has been called the regression function in a textbook I'm using. I'm trying to figure out the relationship between this function and the classical linear regression model.  So, I know that it is a theorem* that we may write  $$ Y = r(X) + \epsilon $$  for some random variable $\epsilon$ s.t. $\mathbb{E}(\epsilon) = 0$.  Now suppose that we have     $$ Y = \beta_0 + \beta_1 X + \epsilon $$   This is the classical 1-dimensional regression function (assuming the $\beta_0$ and $\beta_1$ minimize the residual sum of squares).  Question: Is it then a mathematical theorem that if $Y$ is defined as above, that   $$ r(X) = \mathbb{E}(Y \mid X) = (\beta_0 + \beta_1 X)? $$  And is this why the function $\mathbb{E}(Y \mid X)$ is called the "regression function"?  EDIT: The theorem that I am making use of is as follows (from All of Statistics pg. 89):     Regression models are sometimes written as      $$ Y = r(X) + \epsilon $$      where $\mathbb{E}(\epsilon) = 0$. We can always rewrite a regression model this way. To see this, define $\epsilon = Y - r(X)$ and hence $Y = Y + r(X) - r(X) = r(X) + \epsilon$. Moreover, $\mathbb{E}(\epsilon) = \mathbb{E}\mathbb{E}(\epsilon \mid X) = \mathbb{E}(\mathbb{E}(Y - r(X)) \mid X) = \mathbb{E}(\mathbb{E} ( Y \mid X) - r(X)) = \mathbb{E}(r(X) - r(X)) = 0$. I need to prove that:      if a linear operator $\phi : V \rightarrow V$ on a vector space is nilpotent, then its only eigenvalue is $0$.   I know how to prove that this for a nilpotent matrix, but I'm not sure in the case of an operator. How would I be able to relate $\phi$ to a matrix?  Note: A nilpotent operator $\phi$ has been defined as an operator that satisfies $\phi^{n} = 0$ for some $n \geq 1$. Mysterious! Beamer itemize has no bullets after adding a few packages this is my first time using beamer.  I copied a template from the sharelatex page on Beamer  Which works, and \begin{itemize} shows bullets     But after adding a few more packages that I need, all the bullets are gone!    What is going on and how can I fix this?  \documentclass{beamer}  \usepackage[utf8]{inputenc}    \usepackage{amsmath}  \usepackage{amsthm}     %Theorem Environment  \usepackage{amssymb}  \usepackage{multirow}  \usepackage{cancel}  \usepackage{amsfonts}  \usepackage{graphicx}  \usepackage{subfigure}  \usepackage{upgreek}     \usepackage{mathtools}  \usepackage{thmtools}  \usepackage{enumitem}       %Information to be included in the title page: \title{Sample title} \author{Anonymous} \institute{ShareLaTeX} \date{2014}  \begin{document}    \frame{\titlepage}    \begin{frame}  \frametitle{Sample frame title}  \begin{itemize} \item  This is a text in first frame.  \item This is a text in first frame. \item This is a text in first frame.  \end{itemize}  \end{frame}  \end{document} Proving that $\lim_{n \rightarrow\infty} \int_{0}^{\frac{\pi}{2}} \sin(t^n) dt =0$ It is clear that $\lim_{n \rightarrow\infty} \int_{0}^{1} \sin(t^n) dt =0$. (which is not what is to be proved here)  I don't know how to proceed with the remaining part of the integral ie $\lim_{n \rightarrow\infty} \int_{1}^{\frac{\pi}{2}} \sin(t^n) dt$ (I tried substitution and partial integration).  Can anybody give me a hint ? I find it intuitive enough that the radical of $\mathfrak{gl}_n\mathbb F$ is the scalar matrices, but I have trouble finding an easy, but complete proof:  Proof. Let $\mathfrak s$ denote the scalar matrices. Clearly $\mathfrak s\subset\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$. Suppose that $\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$ is generated by more than one element, so that $X\in\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$, but $X\notin\mathfrak s$. We can change basis such that $\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$ is upper-triangular. (The change of basis leaves the scalar matrices invariant.) Then $X$ is upper-triangular.  I want to conclude that there exists a $Y\in\mathfrak{gl}_n\mathbb F$ s.t. $[X,Y]$ is not upper-triangular. How can I see that $Y$ always exists, except by waving hands? Meaning of hash symbol in pin name I'm reading the datasheet of a PSRAM (available ). In the top-right hand corner all the pins are listed. Some of the pin names end with a hash, e.g. LB#, OE#.  What does the # notation mean for pins? What temperature is achieved in focus point by 5000 flat 1x1cm mirrors onto a satellite dish? There is this  ("R5800 Solar Death Ray") where teenager built a 5000 mirror device which concetrates the solar rays which is showing the potency of the mosaic method of concentrating sunlight using elliptical dish. It can fire the wood, boil the water in seconds, melt an aluminum, glass or even burn the whole in concrere or rock.  What is the approximate maximum temperature (could be a range) is achieved by this mirror device in its focus point on sunny day?       The R5800 solar concentrator, replaced by the new R23k, is made from   an ordinary fiberglass satellite dish. It is covered in about 5800   3/8" (~1cm) mirror tiles. When properly aligned, it can generate a   spot the size of a dime with an intensity of 5000 times normal   daylight. This intensity of light is more than enough to melt steel,   vaporize aluminum, boil concrete, turn dirt into lava, and obliterate   any organic material in an instant. It stands at 5'9" and is 42"   across.   Technical details &amp; calculations:     Calculations: Area of dish aperture (elliptical)= (pi x 102cm x   73cm)/4 = 5848 sq cm       Area of focal point (circular)= pi(0.6cm)^2 = 1.14 sq cm      Dish Type= 0.7m x 1m offset focus elliptical.      Concentration Power= 5000x      Output Power Estimate = 560 watts Interpreting spline results is it safe to keep database connections open for long time I have a .net client application which is connected to a remote database. Is it safe to keep a single connection open for the lifetime of the client (hours)? Does the answer hold if I have multiple (10 or 100) clients running? Let f(x) be a  polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$  $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ . And then there are some question related to this. In the solution it is given that according to the given information, $f(x)$ is a polynomial of degree 4 with leading coefficient 3. $f(x)=3(x-2)(x-3)(x+1)(x+6)+(x^2+1)$ Now what i think is that he got the coefficient 3 using the limits and the rest of the terms$[(x-2)(x-3)(x+1)(x+6)]$ from the respective $f(x)$ values from what was given and also the $(x^2+1)$ was added for the remainders, as they satisfy all the values that have been given to us in the question. I just wanted to know if my approach is right? or is there another elegant way to do it?. What if a light clock travels perpendicular to mirrors that make up the clock? Prove that $\lim_{n\to\infty}a_n\le \lim_{n\to\infty}b_n$ In The Winter Soldier, who did Peggy Carter say she married? When Steve Rogers visits Peggy Carter, she has a family photo on her nightstand and says that she married someone after Cap disappeared. Who did she marry? I am running Ubuntu 16.10 on a MacbookPro and keys 49 and 94 are inverted... I resolved this problem with xmodmap, but this won't apply permanently. After some research I found xmodmap is no longuer use by Ubuntu but xkb instead. So I searched in the configuration of xkb and I found in the config file /usr/share/X11/xkb/keycodes/macintosh this lines :   xkb_keycodes "badmap" {     &lt;TLDE&gt; =  94;     &lt;LSGT&gt; =  49; };  xkb_keycodes "goodmap" {     &lt;TLDE&gt; =  49;     &lt;LSGT&gt; =  94; };   This is exactly my problem, this two keys are inverted. So I wondering if I can't change my layout configuration with the param badmap.    The ouput of setxkbmap -query -v 10  Setting verbose level to 10 locale is C Trying to load rules file ./rules/evdev... Trying to load rules file /usr/share/X11/xkb/rules/evdev... Success. Applied rules from evdev: rules:      evdev model:      pc105 layout:     fr,fr,us variant:    mac,oss, Trying to build keymap using the following components: keycodes:   evdev+aliases(azerty) types:      complete compat:     complete symbols:    pc+fr(mac)+fr(oss):2+us:3+inet(evdev) geometry:   pc(pc105) rules:      evdev model:      pc105 layout:     fr,fr,us variant:    mac,oss, Do different databases use different name quote? For example, mysql quote table name using   SELECT * FROM `table_name`;   notice the `   Does other database ever use different char to quote their table name Minecraft Survival Mode How to print all printable ASCII chars in CLI? While trying to create a custom business rule using Apex I got the error:  INVALID_ID_FIELD, SaveAsActivity is not allowed with whatId that is not supported as a task whatId.  ...and found out that it was caused by the method setSaveAsActivity on my code, that was like:  msg.setSaveAsActivity(true); // "msg" is a SingleEmailMessage   What am I doing wrong here? Bezout's identity in $F[x]$ Let $F$ be a field and $F[x]$ a polynomial ring. Let $p(x)$ be an irreducible polynomial.  Show that $\gcd(p(x),q(x)) = 1\Longrightarrow \exists r(x),s(x)$ such that $r(x)p(x)+s(x)q(x) = 1$.  I know the proof for Bezout's identity for , but this proof uses the notion of absolute value, which cannot be applied to a polynomial ring.   I am thinking about a way to prove the statement: Use the Euclidean algorithm to compute $\gcd(p(x),q(x))$ and show that the result is $1$, and somehow go backwards to obtain $r(x)p(x)+s(x)q(x) = 1$, but it is a bit messy, and I'm not yet able to see it completely.  Please offer some help. If you have some other methods, that would be great, too.