Is there any way to clean a triangulated mesh?
Is there a way to "rescue" mesh that has very bad topology?
Background image resize in Ubuntu 18.04.1 LTS Recently updated to 18.04.1 and now I cannot resize my background image from fill screen to center image.  How can I resize the image now that 18 doesnt have the quick resize feature? Why can other users see the files in my home folder? Reference for $\text{Var}(X)\le (b-a)^2/4$ I am not a statistician, but am working a proof for the upper bound of an expression which contains the variance of a variable which obtains its values from a closed interval, [0,1]. I have seen in several places (mostly forums and web pages) the result that $\text{Var}(X)\le \frac{(b-a)^2}{4}$ for $a\le X\le b$. I would like to use, and cite, this result but am having an unusually difficult time finding a solid reference. Could someone here recommend one for me? Inadvertent use of = instead of == It seems that   if (x=y) { .... }   instead of  if (x==y) { ... }    is a root of many evils.  Why don't all compilers mark it as error instead of a configurable warning?  I'm interested in finding out cases where the construct if (x=y) is useful. I installed kubuntu-desktop just to experiment with its interface.  Then I decided to remove and did the traditional apt-get purge and autoremove but it seems elements of kubuntu are still there. For example interface elements, scrollbars, bootscreen etc are still there.  Is there any method I can completely remove without installing a fresh copy of ubuntu? Prove $x \equiv a \pmod{p}$ and $x \equiv a \pmod{q}$ then $x \equiv a\pmod{pq}$ Derive equation for a cantilever in SHM The Wikipedia  has a derivation of the Klein-Gordon equation.  It gets to this step:  $$\sqrt{\textbf{p}^2 c^2 + m^2 c^4} = E$$  and inserts the QM operators to get  $$\left( \sqrt{ (-i \hbar \nabla)^2 c^2 + m^2 c^4 } \right) \psi = i\hbar\frac{\partial}{\partial t} \psi$$  The article then says     This, however, is a cumbersome expression to work with because the differential operator cannot be evaluated while under the square root sign. In addition, this equation, as it stands, is nonlocal.   To fix this, the first equation is squared instead to get  $$\textbf{p}^2 c^2 + m^2 c^4 = E^2$$  after which the QM operators are inserted and the expression is simplified to get  $$-\hbar^2 c^2 \nabla^2 \psi + m^2 c^4 \psi = -\hbar^2 \frac{\partial^2}{\partial t^2} \psi$$  A couple things I don't understand.  First, are the solutions to this differential equation not exactly the same as the solutions to the first differential equation?  Both sides of the starting equation were squared, so it seems to me that regardless of the particular form of the resulting differential equation, both of them should have the exact same set of solutions.  Secondly, why is the first differential equation cumbersome to work with?  It seems like it would in fact be easier to work with, since the operator under the square root could be expanded in terms of a Taylor series and then you have an equation that is first order in time.  And finally, can someone elaborate on what nonlocal means? The linked article on the Wikipedia page didn't entirely help me understand it. How is the passage of time handled in Pokemon? I stopped following the Pokemon anime a while ago, but I know the series has had Ash go through many regions, including retreading at least one of them.  When Ash was about to get his eight badge, Misty mentioned that they'd been on the road for a year at that point.  I saw some recent episodes, and Ash looks nearly identical to when he first started as a trainer.  Normally I'd attribute that to cartoon sitcom logic, where the characters never age; but in a series with continuity and references to earlier seasons, this makes no sense.  What's the official word on a timeline for the Pokemon anime? Creating surface from points in ArcGIS for Desktop? how to create surface from points in ArcGis? I have a points (centroids of the neighbourhood and number of people living in every neighbourhood) and i would like to create surface that would present density of population. Will tcpdump see packets that are being dropped by iptables? DropDown list onchange event and AJAX in MVC Why is ROC AUC equivalent to the probability that two randomly-selected samples are correctly ranked? I found there are two ways to understand what AUC stands for but I couldn't get why these two interpretations are equivalent mathematically.   In the first interpretation, AUC is the area under the ROC curve. Picking points from 0 to 1 as threshold and calculate sensitivity and specificity accordingly. When we plot them against each other, we get ROC curve.  The second one is that the AUC of a classifier is equal to the probability that the classifier will rank a randomly chosen positive example higher than a randomly chosen negative example, i.e. P(score(x+)>score(x−)).  (from ) I'm reading  and I'm trying to understand the proof under "Degeneracy in One Dimension".  Here's what it says:     Considering a one-dimensional quantum system in a potential $V(x)$ with degenerate states $|\psi_1\rangle$ and $|\psi_2\rangle$ corresponding to the same energy eigenvalue $E$, writing the time-independent Schrödinger equation for the system:       $$-\frac{\hbar^2}{2m}\frac{ \partial^2\psi_1}{ \partial x^2} + V\psi_1 =E\psi_1$$       $$-\frac{\hbar^2}{2m}\frac{ \partial^2\psi_2}{ \partial x^2} + V\psi_2 =E\psi_2$$   Multiplying the first equation by $\psi_2$ and the second by $\psi_1$ and subtracting one from the other, we get:       $$\psi_1\frac{d^2}{d x^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1=0$$   Integrating both sides       $$\psi_1\frac{\partial \psi_2}{\partial x}-\psi_2\frac{\partial \psi_1}{\partial x}=constant$$   In case of well-defined and normalizable wave functions, the above constant vanishes, provided both the wave functions vanish at at least one point, and we find: $\psi_1(x)=c\psi_2(x)$ where $c$ is, in general, a complex constant. So, the two degenerate wave functions are not linearly independent. For bound state eigenfunctions, the wave function approaches zero in the limit $x\to\infty$ or $x\to-\infty$, so that the above constant is zero and we have no degeneracy.   Here are the parts I don't understand:   How does integrating $\psi_1\frac{d^2}{d x^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1=0$ yield $\psi_1\frac{\partial \psi_2}{\partial x}-\psi_2\frac{\partial \psi_1}{\partial x}=constant$.  Without having explicit formulas for $\psi_1(x)$ and $\psi_2(x)$, how do we just apparently pull them both out of the integral? I don't understand how the constant in the last equation vanishes just because $\psi_1$ and $\psi_2$ vanish at infinity. Even if it does vanish, how does that imply $\psi_1=c\psi_2$? Detecting EOF in C++ from a file redirected to STDIN Updating .class file in jar Is there any easy way to open the document library files in a new window or popup depending on the file types?  e.g. PDF - In Adobe reader (currently opens in IE)  Image file (currently opens in IE) My textbook states that:  ...The capacitance $C$ depends only on the geometrical configuration (shape, size, separation)of the system of two conductors. [As we shall see, later, it also depends on the nature of the insulator (dielectric) separating the two conductors.]...  Now I wanted to know if these were the only two factors on which the capacitance of a conductor depends. So I tried thinking about it from the basic definition, which is:  Capacitance is the ratio of the change in electric charge of a system to the corresponding change in its electric potential. (Source: )  i.e., $$C= \frac {dQ}{dV}$$ where, $V$ is the potential of the conductor with respect to zero potential at infinity. Now, I can not think of why $C = f(V)$ isn't a possible scenario, assuming that shape size, etc remains constant. So:  Why cannot $C$ be a function of $V$? I am having an access problem to some files and directories that I have tracked down to the group assigned to access them.    When I manually change the name of the group in the properties/permissions menu for a directory from the default setting (MyGroup) to www-data used by my Drupal Website, the directory error messages I get are gone.    There are a large number of files and directories that require this change.  If believe using chmod will change the owner which is set correctly and being new to Ubuntu, I am reluctant to experiment without a better understanding of how owner and groups work in the permissions settings.   What command changes the Group setting for a directory? I need help with a specific command that I am using in a new game that I am creating inside of Minecraft. The command is:  /give @p sign 1 0 {BlockEntityTag:{Text1:"{"text":" [MF2]","color":"gold","clickEvent": {"action":"run_command","value":"tellraw @a [""{"text":"[MF2]  ","color":"gold"},{"selector":"@p","color":"dark_aqua"},{"text":" is now an admin of the server!","color":"white"}]"}}",Text3:"{"text":"Click to join Admin","color":"dark_aqua","clickEvent": {"action":"run_command","value":"scoreboard teams join admin @p"}}"},display: {Name:"Custom Sign"}}   If I take out the /tellraw command then the command works fine, but I need the command in there to announce to the game that a new admin has joined.  With the tellraw command, the error message is the following:  Data tag parsing failed: Unexpected token 't' at: text":" [MF2]","color":"gold","clickEvent":{"action":"run_command","value":"tellraw @a [""{"text":"[MF2] ","color":"gold"}, {"selector":"@p","color":"dark_aqua"},{"text":" is now an admin  of the server!","color":"white"}]"}}",Text3:"{"text":"Click to join Admin","color":"dark_aqua","clickEvent": {"action":"run_command","value":"scoreboard teams join admin @p"}}"   If anyone knows how to fix this, please let me know.