How should I proceed after black screen during Ubuntu upgrade?
My computer boots to a black screen, what options do I have to fix it?
I'm working on the following problem set.  Let $\mathcal{H}$ be a Hilbert space and $A$ and $B$ orthogonal subspaces of $\mathcal{H}$.  Prove or disprove:  1) $A \oplus B$ is closed, then $A$ and $B$ are closed.  2) $A$ and $B$ are closed, then $A \oplus B$ is closed.  I could prove 1) and if my proof is correct, it even holds if $\mathcal{H}$ is just an inner product space.  Unfortunately, I can't manage to prove 2). Since $\mathcal{H}$ is by assumption a Hilbert space and I didn't use that fact to prove 1), I should probably use it here.  It means that $A$ and $B$ are also complete. Given some convergent sequence in $A \oplus B$, I want to show that the limit is also in  $A \oplus B$. Here I'm stuck. I want to use the completeness of $A$ and $B$, but I don't see how to obtain suitable Cauchy sequences.  Can anyone drop me a hint? Or is my approach all wrong? Checking if a key exists in a JavaScript object? Never delete entries? Good idea? Usual? Javascript infamous Loop issue? Let $P$ be a prime. Show that $\exists$ $x \in \mathbb{N}$ such that $f(x) = p$ then $\exists$ $y \in \mathbb{N}$ such that $g(y) = p$ What is given?  $$\text{Let P be a prime}$$ $$\text{Let} \space f(x)= 3x+1$$ $$\text{Let} \space g(x)= 6x+1$$  Show that: If there exists $x \in \mathbb{N}$ such that $f(x) = P $ , then there exists a $y \in \mathbb{N}$ , such that $g(y) = P$  What have I tried?   Given the question says 'show that', that leads me to believe I need to do a proof of sorts. I'm guessing it might be a proof by either strong induction, or just induction (or maybe just a direct proof?).  I don't even know how to set this up If $f(x) =$ $3x + 1 = $ prime...for any integer $x$, then there exists an integer $y$ such that $g(y) = 6y + 1 = $ prime?   Any hints or starting points would be greatly appreciated.   $$\text{Thanks in advance!}$$ The question I want to ask is related to the answer of this question:  .   My aim is to separate my bibliography into two parts : one part for a personnal bibliography, and a second one for the regular bibliography.  To do that, I use the technique of the post I cited :  \documentclass{article} \usepackage[T1]{fontenc} \usepackage[french]{babel} \usepackage[style=alphabetic,maxnames=6,natbib=true]{biblatex} \bibliography{biblatex-examples}  % Variants of each could be added \newcommand{\firstname}{Donald~E.} \newcommand{\lastname}{Knuth}  \DeclareBibliographyCategory{byname}  \DeclareIndexNameFormat{byname}{% Test could be refined   \ifboolexpr{ test {\ifdefstring{\lastname}{#1}}            and test {\ifdefstring{\firstname}{#3}}               } {\addtocategory{byname}{\thefield{entrykey}}} {}}  \AtDataInput{%   \indexnames[byname]{author}}  \begin{document}  \section{Knuth's books}  \begin{refsection}  \nocite{*}  \printbibliography[category=byname,heading=none]  \end{refsection}  \section{Not Knuth's books}  \printbibliography[notcategory=byname,heading=none] \end{document}   The problem is that only the second bibliography appears, and when I remove the refsection, then all the bibtex entries of my input files are printed, separated as they should.  EDIT: the aim is to have all publications (even the ones not cited) by author knuth in the first section and all the other cited publications in the second one. Are PPAs safe to add to my system and what are some "red flags" to watch out for? Subsurf modifier rounding sharp edges I tried to apply a Subdivision Surface on a part of my object. I can't keep the angles sharp, they all turn into rounded angles.  The angle before applying the Subdivision surface...   ...And after Disease transmitted by postage stamps How to select polygons facing camera instead of polygons located in the opposite side I want to select marked face:    however back face gets selected:    Is there an option in blender to only select facing polygons camera ? $f(x+1)-f(x)$ converges $\Rightarrow\frac{f(x)}x$ converges Let $f:\mathbb{R}\to\mathbb{R}$ continuous such that $\lim_{x\to\infty}f(x+1)-f(x)=l$.  How can I prove that $\dfrac{f(x)}x$ converges for $x\to+\infty$ ?  I am just trying to prove the convergence, not that it converges to $l$ (that part be deduced immediately by Stolz-Cesaro and the sequential characterization of limits) Why (...) doesn't spawn a new child process when run in background? Which way should arrows point in a collapsible accordion? Left/down … down/up? I was looking at  on why  and emission lines don't cancel out:      An experiment shining light on the material and looking at the   reflected spectrum will see absorption lines at those frequencies,   because the relaxation of the excited electrons will emit back   radiation all around randomly, whereas the reflected spectrum is at a   specific angle.   However it is not totally convincing to me. The photons that get emitted towards the center of the Sun (by the electrons in higher energy states that absorbed a photon earlier) would get absorbed again when they hit a different electron (and then re-emitted) and would eventually make it out. So why don't absorption and emission lines cancel out in our Sun? Compute $E(X\mid X+Y)$ for independent random variables $X$ and $Y$ with standard exponential distributions Being $X$ and $Y$ independent random variables distributed as $\mathrm{Exponential}(1)$ and let $T=X+Y$ Calculate $E(X|T)$.  This is my attempt at solving this:  Being $Z=\sum_{i=1}^{n}P_{i}$ where $P_{i}$~$\mathrm{Exp}(\lambda)$ then using it's MFG you have that $$M_{P_i}(t)=\frac{\lambda}{\lambda-t}$$ then $$M_{Z}(t)=M_{\sum{P_{i}}}(t)=(M_{P_{1}})...(M_{P_{n}})=(\frac{\lambda}{\lambda-t})^n$$ with this being the MFG of a $\mathrm{Gamma}$ distribution with parameters $(n,\lambda)$  With this result is easy to see that $T$~$\mathrm{Gamma}(2,1)$  Now $E[X|T]=\int xf_{X|T}(x|t)dx$  Then using Bayes' theorem we can make $$f_{X|T}(x|t)=\frac{f_{T|X}(t|x)f_{X}(x)}{f_{T}(t)}$$ then $$f_{T|X}(t|x)=Pr(T=t|X=x)$$ $$=Pr(X+Y=t|X=x)$$ $$=Pr(x+Y=t|X=x)$$ $$=Pr(Y=t-x)=f_{Y}(t-x)$$  Using this result we can now rearrenge $$f_{X|T}(x|t)=\frac{f_{Y}(t-x)f_{X}(x)}{f_{T}(t)}$$ $$=\frac{e^{-(t-x)}.e^{-x}}{\frac{te^{-t}}{\Gamma(2)}}$$ $$=\frac{1}{t}$$ then $$E[X|T]=\int (x)(\frac{1}{t})dx$$  And this where I got stuck because $1_{x\epsilon(0,\infty})$, I made this explanation step-by-step hoping that someone can se easily check where I went wrong. I would aprecciate a step-by-step answer, thanks. The verb let means  “allow”, “permit”, “not prevent or forbid”, “pass, go or come”  and it's used with an object and the bare infinitive.         Are you going to let me drive or not?     Don't let him off the hook.     Before we let our children surf the Internet, they have to do their homework.     She lets the cats out before leaving.   He let us into the house.      The full form let us can be used similarly        Let us know as soon as possible.      Please, let us help you.   They will never let us forget.      None of the above can be contracted; however, when let us is used for making a suggestion; giving self-encouragement; expressing a consequence or plea,  it is often contracted to let's      Let's go out   Let's have a party   Let's see what happens   Let's stand together in this emergency   Let's not forget those who sacrificed their lives     Questions   I believe that let + us is the only instance where this type of contraction occurs. Is there an explanation as to why verbs such as (i) give + us, (ii) get + us or (iii) take + us were not similarly contracted? When was the apostrophe used to replace the omitted letter "u" in let us?  What is the origin of let's? Plotting Population Regression Function with pgfplot I want the make the following population regression function with pgfplot.     My attempted code is  \documentclass[border=5mm]{standalone} \usepackage{pgfplots}   \begin{document}  \begin{tikzpicture}[ % Define Normal Probability Function declare function={             normal(\m,\s)=1/(2*\s*sqrt(pi))*exp(-(x-\m)^2/(2*\s^2));         }        ] \begin{axis}[     no markers   , domain=-3.2:3.2   , samples=100   , ymin=0   , axis lines*=left   , xlabel=     , every axis y label/.style={at=(current axis.above origin),anchor=south}   , every axis x label/.style={at=(current axis.right of origin),anchor=west}   , height=5cm   , width=12cm   , xtick=\empty   , ytick=\empty   , enlargelimits=false   , clip=false   , axis on top   , grid = major   , hide x axis   , hide y axis   ]    \addplot[cyan!50!black, rotate=-45] (x, {normal(0, 1)});  \end{axis}   \end{tikzpicture}  \end{document}     Any help to get this graph will be highly appreciated. Thanks How to solve ~(P → Q) : P & ~Q by natural deduction? I'm trying to solve the following by natural deduction:  ~(P → Q) : P &amp; ~Q  It's a trivial problem if identities are used, as can be seen by the following:   {1}      1.   ~(P → Q)                    Prem. {1}      2.   ~(~P ∨ Q)                   1 Mat. Impl. {1}      3.   ~~P & ~Q                    2 De Morgan's {1}      4.   P & ~Q                      3 DNE   Starting from step 2 of the previous proof, it's also fairly straightforward to reach a conclusion using natural deduction; i.e. starting from ~(~P ∨ Q), which is an identity of ~(P → Q):  ~(~P ∨ Q) :  P &amp; ~Q   {1}      1.   ~(~P ∨ Q)                   Prem. {2}      2.   ~P                          Assum. {2}      3.   ~P ∨ Q                      2 ∨I {1,2}    4.   ~(~P ∨ Q) & (~P ∨ Q)        1,3 &I {1}      5.   P                           2,4 RAA {6}      6.   Q                           Assum. {6}      7.   ~P ∨ Q                      6 ∨I {1,6}    8.   ~(~P ∨ Q) & (~P ∨ Q)        1,7 &I {1}      9.   ~Q                          6,8 RAA {1}      10.  P & ~Q                      5,9 &I   I also found that it's fairly easy to conclude ~Q:   {1}      1.   ~(P → Q)                    Prem. {2}      2.   Q                           Assum. {3}      3.   P                           Assum. {2,3}    4.   P & Q                       2,3 &I {2,3}    5.   Q                           4 &E {2}      6.   P → Q                       3,5 CP {1,2}    7.   ~(P → Q) & (P → Q)          1,6 &I {1}      8.   ~Q                          2,7 RAA ... {1,2}    12.  P & ~Q   In spite of how easy it looks, I can't figure out how to do it. I believe it can be done in 12 steps. Thanks! Can the Eclipse Java formatter be used stand-alone Is there a way to use the formatter that comes with eclipse, outside of eclipse?  I would like to format some java files using my formatter.xml file that I have configured using eclipse.  Does anyone have any code examples that would allow me to do this?  I would also like to use this standalone, so the specific jars that are used would be nice. Trying to make aliases that open the last modified file One thing that I frequently do is edit the most recently modified files, so instead of typing "ls -lr" and then "vim lastfile", I thought I would make some shortcuts in my ~/.bash_profile file:  alias via="vim `ls -rt | tail -1`" alias vib="vim `ls -rt | tail -2 | head -1`" alias vic="vim `ls -rt | tail -3 | head -1`" alias vid="vim `ls -rt | tail -4 | head -1`" alias vie="vim `ls -rt | tail -5 | head -1`"   The problem is that, weirdly enough, these commands don't work.  They open some file that isn't one of the last, or even a file was deleted from the current directory (I wonder if there's some kind of file cache updating issue in the directory.  This occurs on both my local machine and the cluster I work on).  However, if I type vim `ls -rt | tail -1` directly, without using the alias, it works every time.