Brightness up/F2 button stopped working on my 2014 rMBP Sometime in the last week or two, the F2 button stopped working.. F1 (decrease screen brightness) still works, but not F2.  Is there any kind of fix for this, or is the button broken for good?
Keyboard key stuck or not being recognized; how to fix This is an attempt to write a canonical QA for this issue, as per the Meta post:   I expect it to be periodically edited with the goal of becoming a comprehensive information resource.  I have a MacBook Pro that has a key(Shift, Return,   Command, P, etc.) that doesn’t respond when pressed or is stuck in the down position.    It's possible that my favorite beverage was spilled (or some other liquid damage).  It's also possible that I have no idea what happened.  It was working yesterday and suddenly today it's no longer working.  Is there any way to bypass or remap the key?  Can I  have the system ignore it completely?  (This question can also apply to external USB keyboards or the Bluetooth keyboards)
How do you tell if caps lock is on using JavaScript?  One caveat though: I did google it and the best solution I could find was to attach an onkeypress event to every input, then check each time if the letter pressed was uppercase, and if it was, then check if shift was also held down. If it wasn't, therefore caps lock must be on. This feels really dirty and just... wasteful - surely there's a better way than this? Why does the $\sum_{n>1}(\zeta(n)-1)=1?$ While I was looking at the values of the zeta function for the first natural numbers, I noticed that the sum of the values minus $1$, converge to $1$. Better put: $$\sum_{n=2}^{\infty} \left(\zeta(n)-1\right) = 1 $$ Furthermore, if you use only the even numbers for the zeta function, the sum will converge to $\frac{3}{4}$, or $$\sum_{n=1}^{\infty} \left(\zeta(2n)-1\right) = \frac{3}{4}$$  Leaving $$\sum_{n=2}^{\infty} \left(\zeta(2n-1)-1 \right)= \frac{1}{4}$$  This is probably common knowledge among mathematicians, but I couldn't find much about it on the internet. Is there a proof of this or perhaps even a simple explanation why this is so? Let $ f: \mathbb {R} \rightarrow {\mathbb {R}} $ be given by   $$ f (x) = \left \{\begin {matrix} \sin ({\frac {1} {x}}), &amp; \text {if} \phantom {a} x \neq 0; \\  c, &amp; \text {if} \phantom {a} x = 0 \end {matrix} \right. $$ where $ c \in [-1,1]$. For what values ​​of $c$ is there an antiderivative of $f$?  I do not know of a theorem of the form "if and only if" that  it tells me when a function $ f $ has no antiderivative.  Anyone have a suggestion? I have a random sample of Bernoulli random variables $X_1 ... X_N$, where $X_i$ are i.i.d. r.v. and $P(X_i = 1) = p$, and $p$ is an unknown parameter.  Obviously, one can find an estimate for $p$: $\hat{p}:=(X_1+\dots+X_N)/N$.  My question is how can I build a confidence interval for $p$? Can I use the Adwaita theme in Unity without installing GNOME shell? I recently installed Ubuntu 11.10 (new install). I would like to use the Adwaita theme with Unity, but it seems that it is not available in the theme selection dialog. I know I get it there after installing gnome-shell, but is there a way to use Adwaita theme for Unity without installing gnome-shell? $ p(x)\in\mathbb{R[X]} $ is a polynomial of degree $n$ with no real   roots. Show that: $$\int\limits_{-\infty}^{+\infty}\dfrac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \leq n^{3/2}\pi.$$   It's easy to see that the degree of $ p$ has to be even. For $n=2$ this integral is at most $2\pi$. For $n&gt;2$ the maximum value of this integral is obtained when all the imaginary parts of the roots of $p(x)$ tend to $0$, but I couldn't go further.    Any help would be appreciated, thanks.    Edits by David Speyer: It seems very likely now that the optimum bound is $n \pi$, not $n^{3/2} \pi$. As pointed out in the comments below, and further in , this is the value we get in the limit where $p$ has double roots on the real axis.   It seems likely that  moving the roots of $p$ away from the real axis can only make the integral less. Write the roots of $p$ as $a_j \pm i b_j$, so $p(x) = \prod ((x-a_j)^2 + b_j^2)$ and  $$\frac{p'(x)}{p(x)} = \sum \frac{2 (x-a_j)}{(x-a_j)^2 + b_j^2}.\quad (\ast)$$  So making the $b_j$ larger tends to make $p'(x)/p(x)$ smaller, which makes the integral smaller. But this argument is not rigorous, because the terms of $(\ast)$ can have both positive and negative sign, so it could be that making the individual terms closer to $0$ makes the absolute value of $(\ast)$ larger.  I don't see how to beat this issue easily.  Thus, I'm putting up a bounty for proving or disproving   $$ \int_{-\infty}^{+\infty}\dfrac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \leq n\pi.$$ Someone is offering to pay me in a sort of sugar daddy/sugar baby arrangement. They want to make online deposits to me, and they're asking me for my online banking username and password. Is that information alone enough for them to be able to take my money out of my account, or am I safe? I used the inbuilt dataset stackloss here. I used following R code for creating the multiple regression model.  &gt; head(stackloss)   Air.Flow Water.Temp Acid.Conc. stack.loss   1        80         27         89         42   2        80         27         88         37   3        75         25         90         37   4        62         24         87         28   5        62         22         87         18   6        62         23         87         18  &gt; fit &lt;- lm(stack.loss ~ .,data=stackloss) &gt; plot(fit)   And when I plot the model, the four following graphs are displayed.      How to interpret these graphs? What are the significance of these graphs? How to use these graphs? Sudo doesn't work on Aliases I have added alias ll='ls -la' in /etc/bash.bashrc but sudo ll is not working. Is there a way to "source" bash.bashrc? In a related post, another user suggesting creating an alias for sudo with a space after it. Eg. alias sudo='sudo ' but there wasn't any context to this answer. Should I add this to my bashrc? Seems hackish.  What is the correct method for adding an alias for my sudo one-off commands? How to Make Multi User Objects Single User? Do we really need a [source-code] tag? The  tag doesn't appear to be very useful.  It's used in a variety of ways (for reading, writing, showing, transforming, compiling, translating, obfuscating, obtaining, ... source code) on 1747 questions right now, and doesn't appear to be adding any classification value to the site.  Can we get rid of it? How to use find command to search for multiple extensions I can get all jpg images by using:    find . -name "*.jpg"     But how can I add png files to the results as well? Prove the inequality $n! \geq 2^n$ by induction I'm having difficulty solving an exercise in my course.  The question is:      Prove that $n!\geq 2^n$.   We have to do this by induction. I started like this:   The lowest natural number where the assumption is correct is $4$ as: $4! \geq 2^4 \iff 24 \ge 16$. The assumption is: $n! \ge 2^n$.   Now proof for $(n+1)$ which brings me to: $(n+1)! \ge 2^{(n+1)}$  I think I can rewrite it somehow like this:  $$ {(n+1)} \times {n!}  \stackrel{\text{(definition of factorial)}}{\ge} 2^n \times 2 $$  $$ (n+1) \times 2^n \ge 2^n \times 2 $$  Then I think I can eliminate the $2^n$ and have something like this: $n+1 \ge 2$, or $n \ge 1$.  But I think I'm wrong here some where and was hoping somebody has some advice on this. How can I prove the above assumption?  Any help would be appreciated, kind regards. Is changing your passport a good strategy for removing a negative immigration history in Schengen? When certain nationalities of people want to travel in the Schengen zone, they are required to apply for a Schengen visa. This is a straight-forward process for most, but when an applicant has had problems with removals or refusals they may be reluctant to apply because they want to avoid the complexities of serial visa refusals. This reluctance may lead some into wondering how their history can be expunged, and this line of thought inevitably leads to the idea of getting their passport replaced with a new one.  This sounds like a great formula for people who have had prior refusals or removals.  In essence they can become 'reincarnated' as a first-time traveller and hence avoid the unpleasantness of revealing their prior interactions with the Schengen authorities.  The relevant parts of the application form are shown here...    Nothing asks about a prior removal or refusal, and the applicant is left with simple denials for 3 questions in order to present themselves as a first-time traveller.  Assuming the person is happy to set aside the moral issues associated with denying their history, are there any holes to this strategy?  This question is scoped to Schengen visa applications so that it will not be unmanageably broad, but answers from other regimes (particularly the affluent Commonwealth and the USA) are acceptable as long as the main points are relatively consistent with Schengen. How to read value of variable, where the name of variable is the value of another variable I think an example is in order. Consider I have these two environment variables setup (by a Continuous Delivery tool): PACKAGE_VER=1.2.3 and USE_VERSION_FROM=PACKAGE_VER.  What I want now is a script that can't predict the name of the variable pointing out 1.2.3, only USE_VERSION_FROM. And so, I'm looking for a way to get my script to know the version to use (1.2.3) without knowing anything other than that USE_VERSION_FROM points out the name of the variable that holds the version number.  My naïve non-working first attempt:  #!/bin/bash VERSION_TO_USE="${${USE_VERSION_FROM}}" intuition on the fundamental group of $S^1$ I am familiar with the proof that the fundamental group of the unit circle $S^1$ is $\mathbb Z$, yet I couldn't develop intuition for why it is true. For example, why would I fail if I try to find homotopy from a path that circles $S^1$ once to the path that stays at one point all the time? What brakes the continuity of such a homotopy? I was trying to solve a question of an entrance exam. I am completely stuck in the problem. I am not able to find idea how to proceed. Please help me.     Let $A$ be the set of twice continuously differentiable functions on the interval $[0, 1]$ and let $B = \{f \in A : f(0) = f(1) = 0, f'(0) = 2\}$. Find the value of $$\text{min}_{f\in B} \displaystyle \int_0^1 (f''(x))^2dx.$$    I am really sorry for not showing some effort from my side but I can not find ant way to proceed.  Please help me. Thnx in advance. Background  Recently  was implemented. With this feature the amount of questions with new activity is shown in parenthesis at the beginning of the page title, e.g.     (2) Meta Stack Overflow   means that MSO has two questions with new activity since the user last viewed it.   this feature is unrelated to inbox updates. Here is an example to illustrate this:    Even though there are two inbox messages in this screenshot this is not reflected by the page title. The page title is currently only sensitive to questions with new activity.  Suggestion  I suggest that the page title also reflects inbox updates. For the screenshot above this would mean that the page title shows that there are two inbox updates and two questions with new activity.  I am not sure what format is best for showing this. For chat * in the page title is used to show that there are messages directed at you which is similar in meaning to inbox updates. However, * if is placed after the number in parenthesis to denote inbox updates there might be confusion as there is no connection between the number in parenthesis and * like in chat. Thus, one might prefer another format such as one of the following (all are meant to reflect the status of the screenshot above):      * (2) Meta Stack Overflow     2* (2) Meta Stack Overflow     [2] (2) Meta Stack Overflow     2i 2q  Meta Stack Overflow     (2/2)  Meta Stack Overflow     (2|2)  Meta Stack Overflow    For the alternatives without an explicit hint of which number refers to what it might be best to place the number for questions with new activity first as it will in most cases be greater. This will to a greater degree avoid combinations such as "(0/2)" which may be confused zero of two.  An advantage of the suggested feature is that if the user has a browser tab with a Stack Exchange site then they will be instantly notified of inbox updates. They do not need to focus a tab with a Stack Exchange site to know if there are inbox updates, they merely have to check the relevant page titles among their tabs.  A disadvantage of the suggested feature is that it might be experienced as verbose. I suggest that it is implemented such that it is as unobtrusive as possible while still being informative. Think of a situation where a user has   one Meta Stack Overflow tab with two inbox updates relating to Meta Stack Overflow and two questions with new activity, and one Super User tab with one inbox updates relating to Super User and three questions with new activity.   Then the sum of the inbox notification in the page titles should be local to sites and not global. The page titles for the tabs above should be something like the following pair:      2i 2q  Meta Stack Overflow     1i 3q  Super User How does one determine ladder operators systematically? In textbooks, the ladder operators are always defined," and shown to 'raise' the state of a system, but they are never actually derived.  Does one find them simply by trial and error? Or is there a more systematic method/approach to obtain them? I would like to show that  $$\forall n\in\mathbb{N}^*, \quad \sqrt{\frac{n}{n+1}}\notin  \mathbb{Q}$$  I'm interested in more ways of proofing this.  My method :  suppose that $\sqrt{\frac{n}{n+1}}\in  \mathbb{Q}$ then there exist $(p,q)\in\mathbb{Z}\times \mathbb{N}^*$ such that $\sqrt{\frac{n}{n+1}}=\frac{p}{q}$ thus   $$\dfrac{n}{n+1}=\dfrac{p^2}{q^2} \implies nq^2=(n+1)p^2 \implies n(q^2-p^2)=p^2$$ since $p\neq q\implies p^2\neq q^2$ then  $$n=\dfrac{p^2}{(q^2-p^2)}$$  since $n\in \mathbb{N}^*$ then $n\in  \mathbb{Q}$   I'm stuck here and I would like to see different ways to prove $\sqrt{\frac{n}{n+1}}\notin  \mathbb{Q}$