I'm encountering a problem when I am trying to fill in this textbox. It just doesn't works.  This is the code of the textbox:  &lt;div id="input-field" class="input-field editable" type="text" contenteditable="true" data-ph="Type here to send a message" g_editable="true" role="textbox" style="display: inline-block; border: none; height: 2.7375em; padding: 0.3em 0.3em 0.3em 0.6em; width: 1007px; font-weight: bold; font-style: italic; font-family: 'Comic Sans', 'Comic Sans MS', sans-serif; font-size: 200%;"&gt;&lt;/div&gt;   And this is what I have tried:  document.getElementsByClassName("input-field")[0].value = 'lmao';
I need to set the text within a DIV element dynamically. What is the best, browser safe approach? I have prototypejs and scriptaculous available.  &lt;div id="panel"&gt;   &lt;div id="field_name"&gt;TEXT GOES HERE&lt;/div&gt; &lt;/div&gt;   Here's what the function will look like:  function showPanel(fieldName) {   var fieldNameElement = document.getElementById('field_name');   //Make replacement here }
Since a while, 2 weeks, Dropbox does not startup anymore on my laptop running Ubuntu 12.04 LTS with GNOME shell.  When I run dropbox start -i in the console I get the following output.    Starting Dropbox...Traceback (most recent call last):   File "/usr/bin/dropbox", line 1387, in &lt;module&gt;     ret = main(sys.argv)   File "/usr/bin/dropbox", line 1376, in main     result = commands[argv[i]](argv[i+1:])   File "/usr/bin/dropbox", line 1247, in start     if not start_dropbox():   File "/usr/bin/dropbox", line 731, in start_dropbox     stderr=sys.stderr, stdout=f, close_fds=True)   File "/usr/lib/python2.7/subprocess.py", line 679, in __init__     errread, errwrite)   File "/usr/lib/python2.7/subprocess.py", line 1249, in _execute_child     raise child_exception   OSError: [Errno 8] Exec format error   I already removed Dropbox and reinstalled it. What should I do to fix this? When are "because", "since","for" and "as" interchangeable? I am not a native speaker. On a previous question of mine, I thanked for an answer by saying:     So the phrase is not an idiom, since it is applied in its literal sense.   I consciously chose since over because, because — well, I like that usage of since, even though I know that some (native speakers) don't like it. Nonetheless, that choice got me thinking for a while, and it occurred to me that I even could have used more alternatives, namely as or for, without changing the meaning.   So, on a general note, here are the alternatives:        Statement A (is true), because statement B (is true).   Statement A (is true), since statement B (is true).   Statement A (is true), as statement B (is true).   Statement A (is true), for statement B (is true).      Are these completely interchangeable? Or interchangeable in certain situations? Or can you point out any (not-so-)subtle differences between any of them? Are there even more synonyms? Simple approximation of Inverse Incomplete gamma function In chapter 1, of Landau-Lifshitz Mechanics' book, Landau through isotropy and homogeneity of space and homogeneity of time proves that the Lagrangian must depend of magnitude of velocity of the particle.  This seemed fine. But he didn't give a reason as to why it must depend of $|v|^2$ and not on some $|v|^n$ where $n\neq2$. Is there a function $f\colon\mathbb{R}\to\mathbb{R}$ such that every non-empty open interval is mapped onto $\mathbb{R}$? I wonder  whether there is a function $f\colon\Bbb R\to\Bbb R$ with the folowing characteristic?   for every two real numbers $\alpha,\beta,\alpha\lt\beta$,  $$\{f(x):x\in(\alpha,\beta)\}=\Bbb R$$   I can't  say such a function does not exist, neither can I construct a example  Thanks a lot! Show that the set of polynomials with rational coefficients is countable. I found "A sledgehammer to crack a nut" as one example. What are some others? Useless math that became useful I'm writing an article on  and some people pointed out that this is completely useless.  My idea is to amend my article with some theories that seemed useless when they are created but found use after some time.  I came with some ideas like the Turing machine but I think I'm not grasping the right examples.  Can someone point me some theories that seemed like the Lychrel numbers and then become 'useful'? Lat/Long co-ordinates are not plotting onto EPSG:3857 OSM basemap correctly Excuse me if this is a stupid question, but I've honestly tried to find the answer in a few threads here on SE, and I cannot seem to follow the suggested steps correctly.    Project is in  (WGS 84 / Pseudo-Mercator) BaseMap (OpenStreetMap) layer is set to EPSG:3857 Vector data has been imported from CSV, and has Lat/Long coordinates, set to EPSG:3857.    The layer appears right in the middle of the map, around 0,0 obviously.   Why is this, what am I doing wrong? When to use "have" and "have got" Modified Hamilton's Principle overconstraining a system by imposing too many boundary conditions In Hamiltonian Mechanics, a version of Hamilton's principle is shown to evolve a system according to the same equations of motion as the Lagrangian, and therefore Newtonian formalism. In particular, letting $\delta$ indicate a variation of the path through phase space, $$ \delta \int_{t_1}^{t_2} \big(p_i \dot{q}_i - H(p_i,q_i,t)\big)dt = 0 $$ is shown to generate the same motion equations as those found by the Legendre transform of the Lagrangian equations of motion. Essentially, when we calculate the Euler-Lagrange equations for the above integrand, we find $$ \dot{p}_i + \frac{\partial H}{\partial q_i} = 0\text{, } \dot{q}_i - \frac{\partial H}{\partial p_i} = 0. $$ Now, since these are both first-order equations of motion, we require $2n$ boundary conditions to obtain a solution, where $n$ is the number of particles. This is perfectly consistent with the Lagrangian formalism which had $2n$ boundary conditions due to its $n$ distinct second-order motion equations.   An important thing to note is that since $p_i\dot{q}_i - H(p_i,q_i,t)$ contains no $\dot{p}_i$ dependence, the variation in $p_i(t)$ need not be zero at the endpoints of the path. This is not the case with $q_i(t)$ because the $\dot{q}_i$ dependance results in the appearance of boundary terms that must be set to zero in order to obtain the motion equations.  Now, as far as the boundary conditions are concerned, this makes sense. For the path to be uniquely specified given Newtonian mechanics, we require $2n$ boundary conditions. These can either be the usual Newtonian choice of initial positions and velocities/momenta, or they can be the initial and final positions. Both are perfectly acceptable mathematically speaking. If we were required, in the modified Hamilton's principle variation, to enforce that the variations in the momenta also be zero at the initial and final times, this would mean $2n$ additional boundary conditions, which would generally over constrain a Newtonian system.   However, my confusion is this: when textbooks (Goldstein in particular) consider generating functions, they require that the integrand be invariant to the addition of a total time derivative of a function $F(q_i,p_i,t)$ of the phase space coordinates. However, adding such a function will in general add some $\dot{p}_i$ dependence to the integrand, which will then add boundary terms to the equations of motion unless we require that the $p_i(t)$ variations be zero at the boundaries. This is fine as far as it goes. We can always define the variation however we like. The point is just that it spits out the right motion equations at the end of the day. But my confusion lies in how this doesn't generally over constrain the system. If we specify all the positions and momenta at both the initial and the final times, might not the required path to connect those points in phase space be non-Newtonian? Does this only matter if you're actually trying to use the stationary action principle to find the paths, rather than just using it to find the motion equations? scanf() misbehaving Looking for a Simple Argument for "Integral Curve Starting at A Singular Point is Constant" Let $U$ be an open subset of $\mathbf R^n$ and $V:U\to\mathbf R^n$ be a differentiable vector field on $U$. Let $\mathbf p\in U$ be a singular point of $V$, that is, $V(\mathbf p)=\mathbf 0$.     Then the only integral curve which starts at $\mathbf p$ is the constant curve.   I know that one could simply use the theorem of "uniqueness of integral curves" and in fact adapt the same proof for this particular case.  But is anybody aware of a simple argument for this? What do you call a person who "shuts down" ideas? I really like the , where certain questions from the Stack Exchange Network are presented, along with a hotness rating that is described as &quot;arbitrary&quot; in its tooltip. Such questions also appear randomly on the sidebar on sites across the network, under the heading &quot;Hot Network Questions&quot;. How do these arbitrary hotness points work, and how are they used to select questions to appear in those lists? Weird sharp vertices with subdivision modifier I have a simple skull model, to which I applied the subdivision modifier. For the most part everything works as expected, except I keep getting these weird sharp vertices and I can't seem to figure out how to flatten them.    I'm using Blender 2.8. Is this some kind of feature I can disable/enable for particular vertices? The distance between the sharp vertices and the gray outline differ.    Any suggestions please? Thank you!  EDIT: I also tried selecting as both - vertexes and edges and then applied Edge->Clear Sharp, but nothing changed. Disable password request from suspend - 18.04 matlab based program where output comprises of sum of rows of input I can't use the terminal while gedit command is running I recently installed 12.04. When I try to edit a file with gedit, I can't use the terminal until I close the editing file or I have to open a new terminal. But I think I didn't have this problem with 11.04, however I'm not sure. Is there anyway to avoid this and to use same terminal while editing files. The divisibility rule for $3$ is well-known: if you add up the digits of $n$ and the sum is divisible by $3$, then $n$ is divisible by three.  This is quite helpful for determining if really large numbers are multiples of three, because we can recursively apply this rule:  $$1212582439 \rightarrow 37 \rightarrow 10\rightarrow 1 \implies 3\not\mid 1212582439$$ $$124524 \rightarrow 18 \rightarrow 9 \implies 3\mid 124524$$  This works for as many numbers as I've tried.  However, I'm not sure how this may be proven.  Thus, my question is:     Given a positive integer $n$ and that $3\mid\text{(the sum of the digits of $n$)}$, how may we prove that $3\mid n$?