what is the difference between a = a + b and a += b?
Difference between a += 10 and a = a + 10 in java?
I'm looking for the noun form of "person with intermediate skill". For example, in the context of a particular activity, "person with no skill" might be designated a novice, and "person with much skill" might be designated an expert. However I know of no such word in between these two extremes.   My only thought is amateur, which has a distinct meaning from describing level of skill, although skill level can often be inferred. I'm looking for a better solution.  EDIT: due to the attention this question has received, I'll try to offer some clarification in order to reduce repeat questions.   "Journeyman" would certainly not work. I am a software developer by profession. However the word I'm looking for should hopefully be profession-agnostic. I am trying to designate the skill level of the person that some learning material might be appropriate for. The word that I'm looking for follows the preposition "for".   For Beginners  For [word for intermediate-skilled person] For Experts  Sorry, but even after all the great answers, I'm still learning towards "amateur" even though it is technically incorrect. Also, the word should not be so esoteric that only people with English degrees know what it means. Keep the ideas coming! Diagonal straight lines in residuals vs fitted values plot for multiple regression I am observing strange patterns in residuals for my data:   [EDIT] Here are the partial regression plots for the two variables:      [EDIT2] Added the PP Plot   The distribution  seems to be doing fine (see below) but I have no clue where this straight line might be coming from. Any ideas?   [UPDATE 31.07]  It turns out you were absolutely right, I had cases where the retweet count was indeed 0 and these ~ 15 cases resulted in those strange residual patterns.   The residuals look much better now:   I've also included the partial regressions with a loess line. What is the speed of time? How to properly render smoke on a transparent background? When rendering volumes with Film > Transparent enabled, the alpha appears to be the wrong type (premultiplied/straight). In the 3D view it works as expected:    This is a known .  Is there any way to work around this? Software recommendations Multiple Apple Developer Accounts How to override Drupal.ajax.prototype.beforeSend? I need to add my own little if's and checks in 3 functions inside   misc/ajax.js (   Drupal.ajax.prototype.beforeSend,   Drupal.ajax.prototype.success,   Drupal.ajax.prototype.error )   How can I override them in my module? I can't find a manual and it's not in Examples for Developers. Sorry if it seems a dumb question, but I'm a php developer - I can work with js, but I'm not so fluent with it.   Can I simply write it in my module's js file? If 2 or 3 modules needs to modify them, how to avoid conflicts? Can I remove inline event handlers using jQuery? Proving by induction that $ \sum_{k=0}^n{n \choose k} = 2^n$ Scale and Rotate Stopped Working [2.8] Bypass BIOS password set by faulty Toshiba firmware on Satellite A55-S1065 laptop? Systemd backed tmpfs | How to specify /tmp size manually I mount /tmp on tmpfs using:  sudo systemctl enable tmp.mount sudo systemctl start tmp.mount   But this way /tmp takes up all the free RAM:  $ df -h /tmp Filesystem      Size  Used Avail Use% Mounted on tmpfs           3.9G   12K  3.9G   1% /tmp $   How do I tell systemd tmp.mount to use only 1G? I know I can alternatively not use systemd and manually add an entry to /etc/fstab and specify the size there. But I don't want to do that. I want to use systemd backed tmpfs. Prove a differentiable function between banach spaces is continuous Why do we use decibels instead of just using intensity to measure how loud things are? E:Error, pkgProblemResolver::Resolve generated breaks, this may be caused by held packages I am not sure what caused this error, but here is what the whole error says, and also this is sitting as a notification, and preventing me from updating any software using Update Manager - Please provide some assistance or tell me how to figure out what to do to fix it.  Could not calculate the upgrade  An unresolvable problem occurred while calculating the upgrade.  Please report this bug against the 'update-manager' package and include the following error message: 'E:Error, pkgProblemResolver::Resolve generated breaks, this may be caused by held packages.' I need to compile between two and four times before I see changes I make in the editor (TeXstudio) and each time I compile, a new .tex file is created with random letters and numbers at the end.   For example, If I create the file hello.tex and compile two to four times I'll get the file,   hello.tex.QU6664      then I make some small changes in the editor and compile two to four times I'll get the additional file,   hello.tex.Ak6664   After continuing to do this (each time compiling between two to four times before I see the changes I made), I now have many files that look like  hello.tex.QU6664, hello.tex.Ak6664, hello.tex.YX6664, hello.tex.sN6664, hello.tex.Ss6664, ...   A little bit of additional information (in case it's relevant): Everything was working fine up until a few days ago, i.e. only needed to compile once and only the file hello.tex was created. I still only get one .aux,  .synctex and .pdf files. I use TeXstudio to write and compile. I don't think it has anything to do with the bibliography (I tried removing bibliography altogether and still having the same problem).  How do I prevent these additional files from appearing and only compiling once like I used to? Has anyone had a similar problem? Accessing raster attribute table in QGIS? I have some Python code that makes Raster Attribute Tables for GeoTiffs using GDAL.    In ArcGIS I can see these attribute tables fine and they behave how I like (as classifications) but in QGIS the whole concept of an attribute table seems to be missing.    Is there any way to access a RAT in QGIS? Replace preg_replace() e modifier with preg_replace_callback I stumbled upon  on last weekend, it introduces an integral trick that exploits . On its last page, the author, Mr. Anonymous, left several exercises without any hints, one of them is to evaluate the Gaussian integral $$ \int^\infty_0 e^{-x^2} \,dx= \frac{\sqrt{\pi}}{2} $$ using this parametrization trick. I had been evaluating it through trial and error using different paramatrizations, but no luck so far.     Here are what I have tried so far:   A first instinct would be do something like:$$ I(b) = \int^\infty_0 e^{-f(b)x^2}\,dx $$ for some permissible function $f(\cdot)$, differentiating it will lead to a simple solvable ode: $$ \frac{I'(b)}{I(b)} = -\frac{f'(b)}{2f(b)} $$ which gives: $$ I(b) = \frac{C}{\sqrt{f(b)}}. $$ However, finding this constant $C$ basically is equivalent to evaluating the original integral, we are stuck here without leaving this parametrization trick framework. A second try involves an exercise on the same page: $$ I(b) = \int^\infty_0 e^{-\frac{b^2}{x^2}-x^2}dx. $$ Taking derivative and rescaling the integral using change of variable we have: $$ I'(b) = -2I(b). $$ This gives us another impossible to solve constant $C$ in: $$ I(b) = C e^{-2b} $$ without leaving this framework yet again. The third try is trying modify Américo Tavares's answer in this : $$ I(b) = \int^\infty_0 be^{-b^2x^2}\,dx. $$ It is easy to show that: $$ I'(b) = \int^\infty_0 e^{-b^2x^2}\,dx - \int^\infty_0 2b^2 x^2 e^{-b^2x^2}\,dx = 0 $$ by an integration by parts identity: $$ \int^\infty_0 x^2 e^{- c x^2}\,dx = \frac{1}{2c}\int^\infty_0 e^{- c x^2}\,dx . $$ Then $I(b) = C$, ouch, stuck again at this constant.     Notice in that  question, Bryan Yocks's answer is somewhat similar to the idea of parametrization, however he has to introduce another parametric integration to produce a definite integral leading to $\arctan$.  Is there such a one shot parametrization trick solution like the author Anonymous claimed to be "creative parameterizations and a dose of differentiation under the integral"? This is what we are taught: $$5^{-2} = \left({\frac{1}{5}}\right)^{2}$$  but I don't understand why we take the inverse of the base when we have a negative exponent.  Can anyone explain why?