Can I align some text automaticaly by defined width like this, use only text string as input?   For example, using text "Theoretical bases of the organization of work with documents", I can produce this:    using code:  \parbox{\textwidth-25mm}{ \bf Theoretical \hfill bases \hfill of \hfill the \hfill organization \hfill of \hfill work \hfill with \hfill documents  }   But I needed to manually place \hfill. Can I write a command or a macro to reproduce this result like this.  \somecommand{Theoretical bases of the organization of work with documents}
Is there a way to fully stretch one line of text on the full width of the page, without manually setting the letter spacing, no matter how many characters on the line?
Two similar limit problems I have two limit problems which are quite similar, so I've put them both into this post.   $((4^{10}+2^{n})^{\frac{1}{n}})$ $((3n^{2}+n)^{\frac{1}{n}})$   Attempt:  For 1, I'm not sure if this is permissible. I know that $\lim_{n \to \infty}(x^{n}+y^{n})^{\frac{1}{n}} = \max\{x,y\}$. Obviously $4^{10}$ is a constant so it's not quite in the same form, but I believe it is still okay to use this result, and so the limit is $2$.  For 2, I've got $((3n^{2}+n)^{\frac{1}{n}}) = (n(3n+1)^{\frac{1}{n}})$, but then I'm not too sure how to proceed.   I also thought of taking the log and proceeding that way, however in the notes I'm learning from they haven't covered l'hopital yet, so I'm trying to not use it.  Thanks. Cold fusion is being mentioned a lot lately because of some new setup that apparently works. This is an unverified claim.  See for example:         ( of that last link in the Wayback Machine, given frequent  from that page.)   While we should give the scientific community time to evaluate the set up and eventually replicate the results, there is undoubtedly some skepticism that cold fusion would work at all, because the claim is quite extraordinary.  In the past, after Fleischmann and Pons announced their cold fusion results, in perfectly good faith, they were proven wrong by subsequent experiments.  What are the experimental realities that make Fleischmann and Pons style cold fusions experiments easy to get wrong?  Would the same risks apply to this new set up? How to open a file manager of the current directory in the terminal? When I work in terminal, sometimes I want to open the current directory in a GUI file manager. And then to click the items in the window to run the application. How can I do this? pkexec command in a .desktop file I made a .desktop file for Androxyde's Flashtool (utility for Sony Xperia devices that I have to open with an executable file in its folder) that requires root privileges to use fastboot utilities. I used to make it work with gksu, but I'm on Ubuntu 15.04 and gksu is now old.  I tried to modify the exec line from Exec=gksu /home/natasha/FlashTool/FlashTool to Exec=pkexec /home/natasha/FlashTool/FlashTool Then, I read about pkexec doesn't allow to run X11 applications and so I override in this way:      The problem now is: It asks me the password but the Flashtool's GUI doesn't start. BUT if I execute that command in terminal, the program starts without problems. What can I do? As I understand these two statements:   An electromagnetic field gives particles charge A photon is a quantum of electromagnetic field   It must mean that a photon carries charge. But I guess it isn't true. Why not? Remove trash icon from GNOME desktop and add to Ubuntu dock I just installed Ubuntu 18.04 and don't like having the trash icon on the desktop rather that in the launcher (Ubuntu dock).   How can I move it back to the launcher? Can other icons be moved from the desktop to the launcher? Is it true that one should not use NSLog() on production code? I was told this a few times in this very site, but I wanted to make sure this is really the case.  I was expecting to be able to sprinkle NSLog function calls throughout my code, and that Xcode/gcc would automatically strip those calls out when building my Release/Distribution builds.  Should I avoid using this? If so, what alternatives are most common between experienced Objective-C programmers? Man is sitting in his room and gets contradictory advice from two men from the future Can there be a function that's even and odd at the same time? Is Dirichlet function Riemann integrable? Under what assumptions can a semigroup $(S,*)$ be embedded into a group? Why doesn't the auxiliary will qualify as future tense? According to A Student's Introduction to English Grammar, Tense is defined as follows:     Tense. A system marked by verb inflection or auxiliaries whose basic use to locate the situation in time: I liked it (past tense, past time), I like it(present tense, present time).   However, the author also points out:     ..., English is not one of them: it has no future tense. It does have several ways of talking about future time, and the most basic one does involve the auxiliary will.   These two quotes made me puzzled because, as noted above, Tense is denoted by inflection or auxiliaries and the auxiliary will seems to clearly meet the criteria. In conclusion, my question is   Why doesn't the auxiliary will qualify as future tense? I have two arbitrary lines in 3D space, and I want to find the distance between them, as well as the two points on these lines that are closest to each other.  Naturally, this only concerns the skew case, since the parallel and intersecting cases are trivial.  I know how to find the distance, as the question was asked before and answered .  I haven't found a good explanation on how to find the two points that determine that distance, though.  So specifically, given two lines  $$L_1=P_1+t_1V_1$$ $$L_2=P_2+t_2V_2$$  I would like to find two points $X_1$ on $L_1$ and $X_2$ on $L_2$ such that the distance between $X_1$ and $X_2$ is minimal. 100 prisoners are imprisoned in solitary cells. Each cell is windowless and soundproof. There's a central living room with one light bulb; the bulb is initially off. No prisoner can see the light bulb from his or her own cell. Each day, the warden picks a prisoner equally at random, and that prisoner visits the central living room; at the end of the day the prisoner is returned to his cell. While in the living room, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting the claim that all 100 prisoners have been to the living room. If this assertion is false (that is, some prisoners still haven't been to the living room), all 100 prisoners will be shot for their stupidity. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world can always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity.  Before this whole procedure begins, the prisoners are allowed to get together in the courtyard to discuss a plan. What is the optimal plan they can agree on, so that eventually, someone will make a correct assertion?    I was wondering does anyone know the solution for the 4,000 days solution. Also, anyone got any ideas on how to solve this?   Just wanted to know if anyone got any ideas how to solve this.  is offering prize money for a proof of a optimal solution for it. I suppose to make it clear the people want a plan that would mean on average it would take least amount of time to free the prisoner. The point is the average run time is as low as possible. Maximize $\mathrm{tr}(Q^TCQ)$ subject to $Q^TQ=I$ Let $C \in \mathbb{R}^{d \times d}$ be symmetric, and   $$Q = \begin{bmatrix}       \vert &amp; \vert &amp;    &amp; \vert \\       q_1   &amp; q_2   &amp; \dots &amp; q_K  \\       \vert &amp; \vert &amp;    &amp; \vert       \end{bmatrix} \in \mathbb{R}^{d\times K}$$  where $d \geq K$. Using Lagrange multipliers,   $$\begin{array}{ll} \text{maximize} &amp; \mbox{tr} \left( Q^T C Q \right)\\ \text{subject to} &amp; Q^T Q = I\end{array}$$    I am unfamiliar with these kind of constraints with this method, and after reading  I believe the same specific and simple result given is also applicable, and therefore the lagrangian would be:  $$\mathcal{L}(Q,\lambda)=\mathrm{tr}(Q^TCQ)-\left&lt;\lambda,Q^TQ-I\right&gt;$$  where $\lambda\in\mathbb{R}^{K\times K}$, and $\left&lt;\cdot,\cdot\right&gt;$ is the element wise inner product (what kind of makes sense to me since we're actually adding as many constraints as there are elements in these matrices.  In attempting to do that I start taking $\frac{\partial \mathcal{L}}{\partial Q}=O\in\mathbb{R}^{d\times K}$, and compute that LHS element by element; for the $(l,m)$ one:  \begin{equation} 0=\frac{\partial \mathcal{L}}{\partial Q_{lm}}=(CQ+C^TQ)_{lm}-\underbrace{\frac{\partial}{\partial Q_{lm}}\sum_{i,j}\lambda_{i,j}(Q^TQ-I)_{ij}}_{=\lambda_{lm}\frac{\partial (Q^TQ)_{lm}}{\partial Q_{lm}}}=2(CQ)_{lm}-\lambda_{lm}\frac{\partial (q_l^Tq_m)}{\partial q_m(l)} \tag{1}\end{equation}  where in the last step I've used the definition I made at the beginning for $Q$, and $q_m(l)$ denotes the $l$-th component of the column vector $q_m$.  In trying to compute the very last term:  $$\frac{\partial (q_l^Tq_m)}{\partial q_m(l)}=\frac{\partial \left[q_l(1)q_m(1)+ \ldots + q_l(d)q_m(d)\right]}{\partial q_m(l)}= \begin{cases}     q_l(l)\equiv Q_{ll}    &amp; \text{if } l\neq m\\     2q_l(l)\equiv 2Q_{ll}  &amp; \text{if} l=m \end{cases}$$  The whole equality (1) then can be written:  $$0=2(CQ)_{lm}-\lambda_{lm}Q_{ll}(1+\delta_{lm})$$  where $\delta_{lm}$ is the Kronecker delta.  The equation for the other stationary point of the lagrangian, $\frac{\partial \mathcal{L}}{\partial \lambda}=O\in\mathbb{R}^{K\times K}$, for the $(l,m)$ element as well:  $$ 0=\frac{\partial \mathcal L}{\partial \lambda_{lm}}= \frac{\partial }{\partial \lambda_{lm}}\sum_{i,j}\lambda_{i,j}(Q^TQ-I)_{ij}=(Q^TQ-I)_{lm}\tag{2}$$  what obviously leads to $(Q^TQ)_{lm}=\delta_{lm}$.  All this should tell that the columns of $Q$ are eventually the $K$ first eigenvectors of $C$, but I don't know how to continue from here to prove that, supposing I didn't make a mistake. Please I would sincerely appreciate any help.    Edit:  I have rewritten the inner product as a trace of a product of matrices (after seeing ):  $$\left&lt;\lambda,Q^TQ-I\right&gt;=\sum_{i,j}\lambda_{i,j}(Q^TQ-I)_{ij}=\mathrm{tr}(\lambda^TQ^TQ) $$  and have thus managed to do the derivative without losing the matrix format (using formulas from the ):  \begin{align} O=&amp;\frac{\partial \mathcal{L}}{\partial Q}=\frac{\partial}{\partial Q}\mathrm{tr}(Q^TCQ)-\frac{\partial}{\partial Q}\underbrace{\mathrm{tr}(\lambda^T(Q^TQ-I))}_{\mathrm{tr}(\lambda^TQ^TQ)-\mathrm{tr}(\lambda^T)}\\=&amp;(CQ+C^TQ)-(Q(\lambda^T)^T+Q\lambda^T)=2CQ+Q(\lambda+\lambda^T) \end{align}  And this leads to:  $$CQ=Q\underbrace{\left(-\frac{\lambda+\lambda^T}{2}\right)}_{:=\widetilde{\lambda}};\quad CQ=Q$$  If the defined matrix $\widetilde{\lambda}=Q^TCQ$ were diagonal we would already have the result. Show that an integer of the form $8k + 7$ cannot be written as the sum of three squares. I have figured out a (long, and tedious) way to do it. But I was wondering if there is some sort of direct correlation or another path that I completely missed.  My attempt at the program was as follows:   A number of the form, $8k + 7 = 7 (mod 8)$. That is, we are looking for integers a, b, c such that $a^2 + b^2 + c^2 = 7 (mod 8)$.   LONG and TEDIOUS way: $$(8k)^2 = 0 (mod 8)$$ $$(8k+1)^2 = 1 (mod 8)$$ $$(8k+2)^2 = 4 (mod 8)$$ $$(8k+3)^2 = 1 (mod 8)$$ $$(8k+4)^2 = 0 (mod 8)$$ $$(8k+5)^2 = 1 (mod 8)$$ $$(8k+6)^2 = 4 (mod 8)$$ $$(8k+7)^2 = 1 (mod 8)$$  That is, using three of these modulo there is no way to arrive at  $$a^2 + b^2 + c^2 = 7 (mod 8)$$ How can I integrate $\ln \left( x+\sqrt{1+x^2} \right) $? Change edit approval/rejection I accidentally clicked reject on the suggested edit on this question(). I can't find the button to change it to Approve? I have an Optimus Elite, and it has a minimum brightness setting. In certain apps, I can reduce the brightness below the safe minimum, but when I leave the app, the brightness is increased to the system-wide minimum.  Using , I can permanently set the brightness below the safe amount, and it will stay even after going to other apps. Or using , I can accomplish almost the same thing even without needing root (it displays a translucent 'filter' over the whole screen, reducing brightness/contrast but not actually changing the backlighting power amount).  What I want to know is where the minimum screen brightness is stored. I have a rooted phone with CyanogenMod 9, so I have permissions to edit anything and could change a config or sqlite entry, but I don't know where this value is stored. What I would like to do is use the status bar brightness slider to be able to slide to the actual minimum of the screen, rather than the OS-controlled minimum. Having an app is useful, but having the functionality without the app seems to be within reach, if only I knew where the OS is getting it's safe minimum setting.  Where is this "safe minimum brightness" setting stored? If I can change it, I can remove my apps and use native controls. Will Area 51 be supported in the app?