The following is taken from this book titled "" (Published in 2004):     An independent artist is probably the one who lives closest to an unbounded creative situation. Many artists have considerable freedom from external requirements about what to do, how to do it, when to do it, and why. At the same time, however, we know that artists usually restrict themselves quite forcefully by choice of material and form of expression. To make the choice to express a feeling by carving a specific form from a rock, without the use of high technology or colors, restricts the artist significantly.   In the last sentence, the subject is "To make the choice to express a feeling by carving a specific form from a rock, without the use of high technology or colors," and the predicate is "restricts the artist significantly."  Is it possible to use "Making" instead of "To make" in the last sentence as follows?     Making the choice to express a feeling by carving a specific form from a rock, without the use of high technology or colors, restricts the artist significantly.   Also, how does this gerund subject compare with the to-infinitive subject? Are they both natural in context? Or is one more natural than the other?
In Italian, when I write a sentence about an action (e.g. eating fish, playing), I would use the infinitive, such as  mangiare pesce fa bene alla salute (which literally is "to eat fish is healthy"), or giocare aiuta ad imparare (which literally is "to play helps to learn").   I would use the gerund in giocando si impara, but that would roughly mean "When playing, you learn." and in this case the infinitive would not be correct. (Giocare si impara is not what you would say in Italian.)  Should I use the infinitive, or the gerund, in sentences similar to the following ones?   To play with fire is dangerous. To win the lottery is just a matter of luck. To escape from Alcatraz is impossible. To practice makes you perfect.   Is the infinitive always wrong, when the verb is the subject of a sentence?
Prove that $C\exp(x)$ is the only set of functions for which $f(x) = f'(x)$ I was wondering on the following and I probably know the answer already: NO. Is there another number with similar properties as $e$? So that the derivative of $\exp(x)$ is the same as the function itself. I can guess that it's probably not, because otherwise $e$ wouldn't be that special, but is there any proof of it? How do you become friends with someone? Given that $x^y=y^x$, what could $x$ and $y$ be? It's not too difficult to figure out that $x$ and $y$ can both be 1, and also $x$ can be 2 and $y$ can be 4 (and vice versa). But I can't rule out if there are other solutions. Does it have anything to do with inverse functions? Is there a way to see the solutions by graphing, or algebraically? Shape of a string/chain/cable/rope/wire? The height of a string in a gravitational field in 2-dimensions is bounded by $h(x_0)=h(x_l)=0$ (nails in the wall)  and also $\int_0^l ds= l$. ($h(0)=h(l)=0$, if you take $h$ as a function of arc length) .  What shape does it take?  My try so far: minimise potential energy of the whole string,  $$J(x,h, \dot{h})=\int_0^l gh(x) \rho \frac{ds}{l}=\frac{g \rho }{l}\int_0^l h(x) \sqrt{1+\dot{h}^2} dx$$  With the constraint  $$\int_0^l \sqrt{1+\dot{h}^2} dx- l=0$$ If it helps, it's evident that $\dot{h}(\frac{l}{2})=0$.  Generally, this kind of equation is a case of a constrained variational problem, meaning that the integrand in  $$\int_0^l \frac{g \rho }{l}h(x) \sqrt{1+\dot{h}^2} +\lambda(\int_0^l \sqrt{1+\dot{h}^2} dx- l)dx$$  Must satisfy the Euler Lagrange equation. The constraint must also be satisfied.  But, in truth, by this point I am clueless. $\lambda$ is worked through $\nabla J  = \lambda \nabla(\int_0^l \sqrt{1+\dot{h}^2} dx- l)$. I have tried this , but get nonsensical answers.  Is this method the best? If so, in what ways am I going about it wrongly thusfar? How can I attach a file to a question? How to fix "sudo: unable to open ... Read-only file system"? The title might not be as descriptive as I would like it to be but couldn't come up with a better one.  My server's file system went into Read-only. And I don't understand why it does so and how to solve it.  I can SSH into the server and when trying to start apache2 for example I get the following :  username@srv1:~$ sudo service apache2 start [sudo] password for username: sudo: unable to open /var/lib/sudo/username/1: Read-only file system  * Starting web server apache2                                                                                                                                                                                                               (30)Read-only file system: apache2: could not open error log file /var/log/apache2/error.log. Unable to open logs Action 'start' failed. The Apache error log may have more information.   When I try restarting the server I get :  username@srv1:~$ sudo shutdown -r now [sudo] password for username: sudo: unable to open /var/lib/sudo/username/1: Read-only file system   Once I restart it manually it just start up without any warning or message saying something is wrong.  I hope somebody could point me into the right direction to resolve this issue. Remove "Set as desktop background" from Windows context menu What is the difference between 'git pull' and 'git fetch'? What are the differences between git pull and git fetch? My try, using $x = \sec(u)$ substitution:  $$ \begin{eqnarray} \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x &amp;=&amp; \int \frac{\sqrt{\sec^2(u) - 1}}{\sec(u)}\tan(u)\sec(u) \mathrm{d}u \\ &amp;=&amp; \int \tan^2(u) \mathrm{d}u \\ &amp;=&amp; \tan(u) - u + C \\ &amp;=&amp; \tan(arcsec(x)) - arcsec(x) + C \end{eqnarray} $$  However, according to , the answer should be: $$ \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x = \sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C $$ When I derive this last answer I don't get back the integrand, but rather: $$ \frac{\mathrm d}{\mathrm d x}\left(\sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C\right) = \frac{x}{\sqrt{x^2-1}}- \frac{x}{(x^2-1)^{3/2}\left(1+\frac{1}{x^2-1}\right)} $$  I don't know how to simplify this expression more. Also, I am unable to check whether my answer is correct because I don't know how to find the derivative of $arcsec(x)$.  Can someone check my calculations and tell me where I've done something wrong and how one can simplify the last expression to get back the integrand? Why does pdfTeX hang on this file? How to disable USB storage device support I want to disable a particular USB storage device (a sandisk flash drive). I read somewhere that you can use the .config file. How do i use it to disable that? Usage of “when” or “while” in this context He left the house _ I was still sleeping   Would this sentence be better completed with when or with while? I have a model similar to below with a continuous outcome, continuous predictor and a categorical variable. the goal of this analysis is to make inferences on the group variable. The issue is that y is bounded by x in the form of y >= -x. When the model makes predictions you can see in the plot below that the predicted value is outside the possible range. See the last plot.   How can this be remedied? A custom loss function to penalize going beyond the bound and use gradient descent regression? Or is there some other way? Thank you.   x =   rep(seq(-10,10,1),each=5)  y =  rep(0,length(x) ) weights = sample( seq(1,1000,1) ,length(x)) groups = rep( letters[1:5], times =length(x)/5  )   ## the outcome y has the relationship y &gt;= -x for(i in 1:length(x)){   y[i] = sample(  seq(-x[i],-x[i]+ ifelse(x[i]&lt;0 ,                                            sample(seq(1,10,.1),1),                                           sample(seq(5,40,.1),1) ),1)                        ,1) }   dat =data.frame(x = x, y = y, weights = weights, group = groups) model = lm(dat = dat, y ~ x + group + group * x, weights = weights) dat$FIT= predict(model, dat, allow.new.levels = TRUE)  library(ggplot2) ggplot(data = dat, aes(x = x, y = y, color = group))+geom_point( aes(size = weights))+ geom_line(data = dat, aes(x = x, y = FIT, color = group))    #data for new predictions newdata = data.frame( x = seq(-100,100,1) ,                        group = rep("c", length( seq(-100,100,1) )),                       bound = -seq(-100,100,1)                       )  ## when pridicting new data the outcome y has the relationship y &gt;= -x ## but the predicted vales are below the bound. See how red line is below blue to the left of zero new_preds  = predict(model, newdata, allow.new.levels = TRUE) ggplot(data = newdata, aes(x = x, y = new_preds, color = group))+geom_line()  +   geom_line(data = newdata, aes(x = x, y = bound, color = "green")) what is the relationship between the dynamical casimir effect and virtual particles? Finding empty directories recursively For a continuous function $f$ show $\exists c\in (0,1)$ s.t $f(c)=3c^2$ Let $f: [0,1] \to \mathbb{R}$ be a continuous function such that $$\int_0^1 f (t) dt =1 .$$ Show that there exists a point $c \in (0,1)$ such that $ f(c)=3c^2$.  I tried using mean value theorem of integrals on $f (x)$, so $$\int_0^1 f (t) dt = f (c) =1,$$ but it should be $3 c^2$. What is going wrong? Using $g = \frac{Gm}{r^2}$, the force on a point mass located at 1 AU from the Sun ($m = 2 \cdot 10^{30} \text{ kg}$) is about ~0.006 N/kg.  Does that mean that, e.g., a 70 kg person is ~42g lighter during the day, and ~42g heavier at night? That seems like it could make a big difference for things like measuring gold bars or other weight-sensitive items. (Gold arbitrage: buy your gold during the day and sell it during the night! Risk-free profit!)  This makes me suspect that I'm overlooking something obvious, because a ~0.05% weight difference seems like something everyone would have noticed long ago. So what am I missing?    Edit: A few answers below indicate that there shouldn't be any weight difference, because the Earth orbits the Sun in free-fall. But if that's the reason, does that mean that a 1:1 tidally-locked Earth-Sun system wouldn't experience differential gravity from the Sun on opposite sides? That doesn't seem right. E: Problem executing scripts APT Update::Post-Invoke-Success error during apt-get update I'm trying to install bang and olufsen drivers, or HP hd audio driver, as only 2 of the 4 speakers work and the 2 amps don't work in ubuntu  lspci -v | grep -i audio 00:03.0 Audio device: Intel Corporation Broadwell-U Audio Controller (rev 09)     Subsystem: Hewlett-Packard Company Broadwell-U Audio Controller 00:1b.0 Audio device: Intel Corporation Wildcat Point-LP High Definition Audio Controller (rev 03)     Subsystem: Hewlett-Packard Company Wildcat Point-LP High Definition Audio Controller   The volume is terrible. Is there any way to fix this?  The speakers work perfectly on the pre-installed Windows 8. Is it possible to store user's crontab in user's git repository on OpenBSD? On OpenBSD and with OpenBSD's  and , is it possible to store the  of a user in a  repository of the same user?  What would be the proper way to accomplish something like this?  (To steer the answer into the correct direction, I would not be opposed to changing some permissions in the system, although I'd rather not have to recompile the binaries, nor violate any good security paradigms.)