Can't mount ISO file for MATLAB as loop device: Error: “failed to setup loop device” I want to install Matlab 2018 in Linux. I tried running the following commands:  $ sudo mkdir -p /mnt/cdrom $ sudo mount MatlabR2018b_LinuxX64_DVD1_downloader.ga.iso /mnt/cdrom -o loop   But I was faced with this error:  mount: /mnt/cdrom: failed to setup loop device for MatlabR2018b_LinuxX64_DVD1_downloader.ga.iso.   I have already read and tried what is proposed in this thread but it didn't work:   How can I fix this?
Can't mount ISO file as loop device: Error: “failed to setup loop device” sudo mkdir -p /media/cdrom cd ~ sudo mount -o loop ubuntu-* / mount: ubuntu-*: failed to setup loop device: No such file or directory
Lagrangian of Schrodinger field The usual Schrodinger Lagrangian is $$ \tag 1 i(\psi^{*}\partial_{t}\psi ) + \frac{1}{2m} \psi^{*}(\nabla^2)\psi, $$ which gives the correct equations of motion, with conjugate momentum for $\psi^{*}$ vanishing. This Lagrangian density is not real but differs from a real Lagrangian density  $$ \tag 2 \frac{i}{2}(\psi^{*}\partial_{t}\psi -\psi \partial_{t}\psi^{*} ) + \frac{1}{2m} \psi^{*}(\nabla^2)\psi $$ by a total derivative.  My trouble is that these two lagrangian densities lead to different conjugate momenta and hence when setting equal time commutation relations, I am getting different results (a factor of 2 is causing the problem). I can rescale the fields but then my Hamiltonian also changes. Then applying Heisenberg equation of motion, I don't get the operator Schrodinger equation.   Is it possible to work with the real Lagrangian density and somehow get the correct commutation relations? I would have expected two Lagrangians differing by total derivative terms to give identical commutation relations (since canonical transformations preserve them). But perhaps I am making some very simple error. Unless all conjugate momenta are equivalent for two Lagrangians differing by total derivatives, how does one choose the correct one?   I guess the same thing happens for other first order systems like Dirac Lagrangian also. Let $a_n$ be a sequence of positive reals, such that the partial sums $S_n = \sum_{i = 1} ^ n a_i$ diverge to $\infty$. For given $\epsilon &gt; 0$ do we have $$\sum_{n = 1} ^ \infty \frac{a_n}{S_n^{1 + \epsilon}} &lt; \infty?$$  For $\epsilon \ge 1$ we can resolve this quickly by noting $$\frac{a_n}{S_n ^ 2} \le \frac 1 {S_{n - 1}} - \frac 1 {S_n}$$ so for sufficiently large $n$ we can bound $\frac{a_n}{S_n^{1 + \epsilon}}$ by $\frac 1 {S_{n - 1}} - \frac 1 {S_n}$ as well. I'm wondering if this is true for arbitrary $\epsilon &gt; 0$. I know that the series in question diverges for $ \epsilon = 0$, so all that is missing is what happens in $(0, 1)$. $$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} $$  According to my textbook the limit equals $2$.  What I have tried:  Using the squeeze theorem:  $$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 2} $$ $$ 0 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le 0 $$  I have also tried to use the squeeze theorem with two other equations and obtained different values:  $$ \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 - 1}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 + 1}{\sqrt{x^2 +y^2 + 1} - 2} $$ $$ -1 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le -1 $$ How to import a module given the full path? How to move files to specific folders based on CSV entries? An element of a group has the same order as its inverse If $a$ is a group element, prove that $a$ and $a^{-1}$ have the same order.   I tried doing this by contradiction.    Assume $|a|\neq|a^{-1}|$  Let $a^n=e$ for some $n\in \mathbb{Z}$ and $(a^{-1})^m=e$ for some $m\in \mathbb{Z}$, and we can assume that $m &lt; n$.  Then $e= e*e = (a^n)((a^{-1})^m) = a^{n-m}$. However, $a^{n-m}=e$ implies that $n$ is not the order of $a$, which is a contradiction and $n=m$.  But I realized this doesn't satisfy the condition if $a$ has infinite order. How do I prove that piece? I have a training set with about 3000 positive instances and 3000 negative instances. But my test data set is pretty much un-balanced. The positive set only has 50 instances and negative has 1500 instances.This causes the precision very low. Are there any approaches to solve this problem? I use SVM to build classifier. What is the difference between <%# and <%= in ASP.NET MVC? Eeeek! Something stole the protection! (Why is this protected question unprotected?) asked me to report this (apparent) bug.  When he tried to protect  question someone (or something) stole the question's protection, which has since been missing:          The question shows no sign of having the protection removed, thus I believe it was abducted. I am sure justice will be done, and this poor question will receive the protection it deserves.     I recommend interrogating the  that was said to have announced this question to the world, I bet it would have some useful information. How to restart the networking service? I tried to use  sudo service networking restart   and  sudo /etc/init.d/network restart   but they both crash the window manager and I can no longer use my keyboard for input into X.  when I use the /etc/init.d/ method it complains saying that I should use the service utility  e.g. service networking restart   but it crashes just the same.  Is there a GUI method of restarting networking? Coupon collector's problem: mean and variance in number of coupons to be collected to complete a set (unequal probabilities) There are $n$ coupons in a collection. A collector has the ability to purchase a coupon, but can't choose the coupon he purchases. Instead, the coupon is revealed to be coupon $i$ with probability $p_i=\frac 1 n$. Let $N$ be the number of coupons he'll need to collect before he has at least one coupon of each type. Find the expected value and variance of $N$. Bonus: generalize to the case where the probability of collecting the $j$th coupon is $p_j$ with $\sum\limits_{j=1}^n p_j=1$  I recently came across this problem and came up with/ unearthed various methods to solve it. I'm intending this page as a wiki with various solutions. I'll be posting all the solutions I'm aware of (4 so far) over time.    EDIT: As mentioned in the comments, this question is different from the one people are saying it is a duplicate of since (for one thing) it includes an expression for the variance and it covers the general case where all coupons have unequal probabilities. The case of calculating the variance for the general case of coupons having unequal probabilities has not been covered anywhere on the site apart from , which this one intends to consolidate along with other approaches to solve this problem.    EDIT: Paper on the solutions on this page submitted to ArXiv: How to execute file I'm editing in Vi(m) find an approximate solution, up to the order of epsilon The question is to find an approximate solution, up to the order of epsilon of following problem.  $$y'' + y+\epsilon y^3 = 0$$ $$y(0) = a$$ $$y'(0) = 0$$  I tried to solve the given problem using perturbation theory.  $$y(t) = y_0(t) + \epsilon y_1(t) + \epsilon^2 y_2(t) + \cdots$$ $$t = w(\epsilon)t = (1 + \epsilon w_1 + \epsilon^2 w_2 + \cdots )t$$  However, i failed to find an appropriate approximate solution... help me!! Simulate bad cell phone reception in the lab This is similar to  but I want the iterate tool to run to collect all the values and only then continue to the merge. Any ideas? Currently it just runs the whole process for each file (which works but takes much longer than if the merge is done only once) or I have to manually break the collect-merge link, run the iterate, reconnect the collect-merge link and then continue running the model which is a pain as it is a lot of towers that need processing and this collect values-finish getting all values and then continue will be required in other models as well.    I tried the "Select Data" tool as well but this doesn't allow for wildcard searches like the Iterate does -my Iterate has the "_Graded*" as the search param as the gdb included many other files related to each tower chunk.  My STOP is set to "True" (tried false as well). I also tried to connect the Stop to the Iterate instead of Collect values but same result.  === SCRIPT ADDED  I just created a python tool for this and it collects all the values but it passes them back to the model as a long list -how can I get it to output in a proper format?      import sys, os, arcpy  GDB=arcpy.GetParameterAsText(0) Search=arcpy.GetParameterAsText(1) Type=arcpy.GetParameterAsText(2)  ##GDB=r'P:\2012\044_DM_Radio_Propogation\Working\Propogation\Propogation_28355.gdb' ##Search='_m' ##Type='Polygon'  path=os.path.dirname(GDB) arcpy.env.workspace = GDB List=[] for item in arcpy.ListFeatureClasses('*'+Search+'*',Type):     List.append(path+"\\"+item)     #arcpy.SetParameterAsText(3, item)  arcpy.SetParameterAsText(3, List) arcpy.AddMessage("List: "+str(List)) Is there a dollar amount that, when adding Massachusetts Sales Tax, precisely equals $200? Combine two columns into a list of all possible combinations of entries I have two columns such as:  a   1  b   2  c    and I need to combine them like:  a   1 a   2 b   1 b   2 c   1 c   2   Is it possible with a formula? Keep processes running after SSH session disconnects How to test for equality of means for proportions? And two-sided or one-sided? I have the following problem: two groups A and B.  The proportions of married and single are ($p_1$) and ($p_2$) respectively, and the standard deviations ($s_1$) and ($s_2$). The sample consists of  ($n_1$) A and  ($n_2$) B observations.  Now I want to test whether the proportions are equal. I use the following formula to compute the z-statistic:  z = $abs(p_1 - p_2) / \sqrt{ \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2} }$  and test the hypothesis: h0: $p_1 \neq p_2$  An alternative approach I tried is to estimate the t-value based on the standard deviations (hence just comparing means), that is by the formula (assuming unequal variances):  t = $abs(p_1-p_2)/\sqrt{ s_1^2/n_1 + s_2^2/n_2 }$  My question are:    do you agree with my approach or; should it be two-sided or one-sided?   Thank you. Let thing of a bird standing still in a box on top of a weighing machine that shows a mass $m_0$. Now, imagine that the bird is flying, still in the same box and the same weighing machine shows a mass $m_1$. As the bird when flying, is applying a force towards the weighing machine, could we deduce that $m_0 = m_1$? I'm asking this question because saying that these masses are equal makes me as uncomfortable as saying that they are not equal and can't figure the right answer. So does $m_0 = m_1$ and why?