Changing back the user rights to sudo
How do I add myself back as a sudo user?
Warning: mysql_fetch_* expects parameter 1 to be resource, boolean given error Confusion between Voltage gain & Voltage gain in decibels (dB) The definition of voltage gain is \$V_{\text{out}}/V_{\text{in}}\$.  However, I read some articles about the gain in decibels, and I have a confusion now.  Here is an article about it:   Here, I understand the definition of Power gain in decibels, which is $$ \text{Gain} = 10 \log \left( {P_{\text{out}} \over P_{\text{in}}} \right)\text{ dB} $$  However, I can't understand why Voltage gain in decibels is $$ 20 \log \left( {V_{\text{out}} \over V_{\text{in}}} \right)\text{ dB} $$  If \$ 20 \log \left( {V_{\text{out}} \over V_{\text{in}}} \right)\text{ dB} \$ is derived from $$ 10 \log {\left( {V_{\text{out}}^2 \over R_{\text{out}}}\right) \over \left({V_{\text{in}}^2 \over R_{\text{in}}}\right) }\text{ dB} $$ then this is the power gain, not the voltage gain, isn't it? However, the Wikipedia says it is a formula for the Voltage gain in decibels. I thought the voltage gain in decibels would be \$ 10 \log \left( {V_{\text{out}} \over V_{\text{in}}} \right)\text{ dB} \$. Actually, the example section in that linked page uses voltage gain \$ V_{\text{out}} \over V_{\text{in}} \$.  Why did \$V^2/R\$ suddenly come out from the voltage gain in decibels? "May" & "Might": What's the right context? I may not be coming in tomorrow...    I might not be coming in tomorrow...       When should I use "may" and "might"? In Portal, we are promised cake and a party at the end of our tests. However;       we soon discover that the cake is a lie.     Then, at the final scene;       there really is a cake (and my companion cube is waiting for me).   What's the meaning of this?  Was GLADOS not deceiving me? Her comments seem to suggest that;       ...she had the intention of killing me. Was the cake for us (player) and not for the character in the game? Calculation of $\int^{\pi/2}_{0}\cos^{n}(x)\cos (nx)\,dx$, where $n\in \mathbb{N}$ Compute the definite integral      $$ \int^{\pi/2}_{0}\cos^{n}(x)\cos (nx)\,dx $$      where $n\in \mathbb{N}$.   My Attempt:  Using $\cos (x) = \frac{e^{ix}+e^{-ix}}{2}$, we get  $$ \begin{align} \int^{\pi/2}_{0}\cos^{n}(x)\cos (nx)\,dx&amp;=\int_{0}^{\pi/2} \left(\frac{e^{ix}+e^{-ix}}{2}\right)^n\left(\frac{e^{inx}+e^{-inx}}{2}\right)\,dx\\ &amp;= \frac{1}{2^n}\mathrm{Re}\left\{\int_{0}^{\pi/2} \left(e^{ix}+e^{-ix}\right)^n\cdot e^{inx}\,dx\right\}\\ &amp;=\frac{1}{2^n}\mathrm{Re}\left\{\int_{0}^{\pi/2}\left(e^{2ix}+1\right)^n\,dx\right\} \end{align} $$  Letting $z=e^{4ix}$ gives us  $$ \begin{align} e^{4ix}dx &amp;= \frac{1}{4i}dz\\ dx &amp;= \frac{dz}{4iz} \end{align} $$  So the integral becomes  $$\frac{1}{4\cdot 2^n}\mathrm{Re}\left\{\int_{C}\left(\sqrt{z}+1\right)\cdot \frac{dz}{iz}\right\}$$  How can I complete the solution from here? What laptop specs are important for intensive image editing work? How important is the following in purchasing a laptop that will be used for image editing:   processor speed amount of RAM integrated vs. dedicated graphics card HD speed (SSD; or 5400/7200/10000 rpm) Let $E$ be a Banach space, $\mathcal B(E)$ the Banach space of linear bounded operators and $\mathcal I$ the set of all invertible linear bounded operators from $E$ to $E$. We know that $\mathcal I$ is an open set, and if $E$ is finite dimensional then $\mathcal I$ is dense in $\mathcal B(E)$. It's not true that $\mathcal I$ is dense if we can find $T\in\mathcal B(E)$ injective, non surjective with $T(E)$ closed in $E$, since such an operator cannot be approximated in the norm on $\mathcal B(E)$ by elements of $\mathcal I$ (in particular $E$ has to be infinite dimensional).  So the question is (maybe a little vague): is there a nice characterization of $\overline{\mathcal I}^{\mathcal B(E)}$ when $E$ is infinite dimensional? Is the case of Hilbert space simpler? Data Dump of Attachments I am trying to do an export of approx 30,000 attachments in our Salesforce Org,  Would anyone have been through a similar exercise and can advise best practice or the best way / tool to do this?  Kind Regards  matt Is it "I" or "me" in "Keep Tom and I/me updated"? I am reading Velleman's "How to Prove it", and in one example he proves  $\neg \exists x P(x)  \implies \forall x \neg P(x)$    in the following way:    Suppose $\neg \exists x P(x)$. Let x be arbitrary. Suppose $P(x)$. Since we have a specific $x$ for which P(x) is true, it follows that $\exists x P(x)$, which is a contradiction. Therefore $\neg P(x)$. Since $x$ was arbitrary, we can conclude that $\forall x \neg P(x)$, so $\neg \exists x P(x)  \implies \forall x \neg P(x)$    What makes me confused is the line in bold text, that is "Suppose $P(x)$". The reason is that he writes that he is doing proof by contradiction, but $P(x) \not\iff \neg\forall x \neg P(x)$. He is not treating all possible cases where $\forall x \neg P(x)$ is false by assuming $P(x)$, does he? So how can this be a valid proof by contradiction?    Edit: I deleted my first example since it was clearly not representative for the problem.  Edit 2: So the thing that I am opposing I guess is that $\neg \forall x \neg P(x)$ is not $P(x)$. And from what I have undestood, a proof by contradiction should use the negation of the consequent which I do not think that Velleman does. "World's Hardest Easy Geometry Problem" Alternative to Adobe Illustrator Can you recommend a good alternative to Adobe Illustrator?   I am learning vector based graphic design, but I am not ready to make such a large investment as to purchase a copy of Adobe Illustrator. Also, Illustrator might be a little more power than I need at the moment. I am using a Macintosh. I have an Ubuntu 12.04 machine on which I have installed VirtualBox with Ubuntu 12.04 as the guest OS.  How can I share files between the host and guest?   I have installed the guest additions also but still it’s not able to mount the sf_shared folders if even it’s mounted. This means I'm unable to retrieve the file from the host. Stack exchange badges It might be useful to encourage existing users of stackoverflow to cross pollinate other stack exchange sites.  While rep-sharing is not useful, as rep is meant to be a measure of expertise in one subject, badges may be one way to encourage participation across many sites.  This was triggered by seeing how few people actually had greater than 1k rep on all three main sites (SO, SF, SU) via Dennis Williamson's  app.  I had tried to do this by hand prior to the API under the assumption that many people would be answering questions on all three sites, but it turns out there are only 33 people, of all the users of stackoverflow, that have a 1k+ reputation on all three sites.  Over time it may be useful to provide badges that appear on all sites one is participating in that are awarded for extraordinary participation in the whole universe of Stack Exchange sites.  For instance, a silver badge might appear on all my profiles if I obtain 1k rep on 5 or more sites, and a gold badge if I have more than 3k rep on 5 or more sites.  I would hesitate extending this much past 5 sites as we don't want to encourage spamming questions and answers on sites where the user has little interest, but across the universe of stackapps I know I'm proficient or expert in at least 10 current sites, and while it's not a big nudge, the idea that I can get extra awards - especially ones that appear on all sites I'm in - would move me to participate a little more on sites that I am registered on, but don't visit regularly.    I expect, eventually as sites are launched, most users of any one site will have some interest or knowledge in at least 5 other sites.  This may provide the extra nudge some sites need to maintain an excellent rating in terms of usage. I cannot find enable-interactive-search in    gsettings set org.gnome.nautilus.preferences enable-interactive-search true.     I searched using    gsettings list-recursively | grep search     and    gsettings list-recursively | grep interactive     but no such option.  Can you help me disable this in Ubuntu 17.10 and later? How to delay or slowly loop commands? I found 11 questions like this on gamingSE already, but none of them have proper and up-to-date answers. So I'm creating this Q&amp;A:  How do you either activate a command x seconds/minutes/hours/… in the future or every x seconds/minutes/… in a loop?  Repeating command blocks and ticking functions execute 20 times per second, redstone is  in many situations and most of the extremely fancy solutions, like falling blocks in cobwebs, have broken over time or are needlessly complicated. I've seen in every introductory material on LaTeX and in a lot of code of other people (event experts) that the \newcommand command is used by wrapping the first argument in braces, even if it seems unnecessary to me.   The braces are consistent with the usual way arguments are passed to LaTeX commands (since usually the braces are used even if the argument is made of a single token, for consistency and clarity), but in this single case they appear very ugly to me and in my opinion they lower readability.  But these are opinions, while the question is technical: is there any technical reason why one should write this:  \newcommand{\mycmd}[1]{Hello #1}   instead of this:  \newcommand\mycmd[1]{Hello #1} Where can I see how many posts have I edited? Sum of nth roots of unity Question: If $c\neq 1$ is an $n^{th}$ root of unity then, $1+c+...+c^{n-1} = 0$  Attempt: So I have established that I need to show that $$\sum^{n-1}_{k=0} e^{\frac{i2k\pi}{n}}=\frac{1-e^{\frac{ik2\pi}{n}}}{1-e^{\frac{i2\pi}{n}}} =0$$ by use of the sum of geometric series'. My issue is proceeding further to show that this is indeed true. Confidence bands for QQ line This question doesn't specifically pertain to R, but I   chose to use R to illustrate it.  Consider the code for producing confidence bands round a (normal) qq-line:  library(car) library(MASS) b0&lt;-lm(deaths~.,data=road) qqPlot(b0$resid,pch=16,line="robust")   I'm looking for an explanation of (or alternative a link to a paper/online document explaining) how these confidence bands are constructed (I have  seen a reference to Fox 2002 in the help files of R  but sadly I don't have this book handy).   My question will be made more precise with an example.  Here's how R computes these particular CI's (I've  shortened/simplified the code used in car::qqPlot)  x&lt;-b0$resid good&lt;-!is.na(x) ord&lt;-order(x[good]) ord.x&lt;-x[good][ord] n&lt;-length(ord.x) P&lt;-ppoints(n) z&lt;-qnorm(P) plot(z,ord.x,type="n") coef&lt;-coef(rlm(ord.x~z)) a&lt;-coef[1] b&lt;-coef[2] abline(a,b,col="red",lwd=2) conf&lt;-0.95 zz&lt;-qnorm(1-(1-conf)/2) SE&lt;-(b/dnorm(z))*sqrt(P*(1-P)/n)     #[WHY?] fit.value&lt;-a+b*z upper&lt;-fit.value+zz*SE lower&lt;-fit.value-zz*SE lines(z,upper,lty=2,lwd=2,col="red") lines(z,lower,lty=2,lwd=2,col="red")   The question is: what is the justification for  the formula used to compute these SE (e.g. the line SE&lt;-(b/dnorm(z))*sqrt(P*(1-P)/n)).   FWIW this formula is very different from the formula of the usual