Integer object leading zero changing the value
Why are integer literals with leading zeroes interpreted strangely?
Find the determinant of $I + A$ Highlight elements in the matrix I have a matrix and need to highlight some elements in the matrix. I have found how to highlight a single element (I want to put the cell in the square):  $$ A = \begin{bmatrix} \fbox{0} &amp; \fbox{0} &amp; 0 \\  \fbox{0} &amp; \fbox{0} &amp; 0 \\ 1 &amp; 1 &amp; 1 \\  \end{bmatrix}$$   But how can I outline elements of the submatrix like in the picture below? I need just a frame where some elements are located. Ways to eliminate switch in code When saving a .qgs file in QGIS, another appears. Is this a backup? When saving a project file (.qgs), which when executed is automatically loaded with qgis.bat, there's another .qgs file which appears in the same directory of the original project file, although this one is not loaded with the qgis.bat file:    This is just out of curiosity but is the latter designed to be a backup of the original .qgs file? Why does a free-falling body experience no force despite accelerating? Find out which process is locking a file or folder in Windows Is it possible to remove a terminal command? For example, if there was a command call lookup (not an alias), how would I go about removing it? Should "move-to-chat" be provided on demand (as a button perhaps)? How old is Chewbacca, anyway? Do Wookiees get gray fur when they are old? I just saw a user named: .yahoo.co.jpaqwsykcj3aulh3h1k0cy6nzs3isj  Could new users be named according to their user-number, or something other than "Unknown"? Perhaps User1234567 instead? It would be less awkward to address them in answers and comments.   Additionally, I often times have to check profiles to make sure users aren't abusing the system, since I find many of the same questions from "unknown (google)" from time to time. That, or I'll mistakenly think I've helped the user on a similar issue in the past, and assume the user has knowledge that he or she does not have. This is a batch file in Windows.  Here is my .bat file          @echo off copy "C:\Remoting.config-Training" "C:\Remoting.config"  "C:\ThirdParty.exe"   This works fine except the .bat file leaves the command window open the whole time the "ThirdParty" application is running. I need the command window to close.   I would use the short-cut for the application but I must be able to run this copy command first (it actually changes which data base and server to use for the application).    The ThirdParty application does not allow the user to change the source of the db or the application server.    We're doing this to allow users to change from a test environment to the production environment. Mapping half-plane to unit disk? Say you have the half-plane $\{z\in\mathbb{C}:\Re(z)&gt;0\}$. Is there a rigorous explanation why the transformation $w=\dfrac{z-1}{z+1}$ maps the half plane onto $|w|&lt;1$? I have three displays, a 4K display flanked by two 1080p displays. Physically, these monitors are roughly the same size. Furthermore, Windows 10 handles the DPI scaling of the 4K monitor separately from the 1080p monitors just fine. So while it appears as such to Windows    the real layout is three monitors of equal size side by side.  The problem is that the mouse keeps getting caught on the edges of display 2 and three when dragging from display 1. I would like the mouse to continue onto display 2 or 3 regardless of which part of the left or right edge of display 1 the mouse touches.  This question has been asked before (e.g. ) and a common answer is to suggest that this behaviour is intentional on the basis that mouse movement should reflect different size displays. Such an answer is not appropriate as it doesn't consider DPI and scaling. I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ which are continuous and satisfy $f(x+y)=f(x)+f(y)$ I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $f(x+y)=f(x)+f(y)$. I know that there are other questions that are asking the same thing, but I'm trying to figure this out by myself as best as possible. Here is how I started out:  Try out some cases:  $x=0:$ $$f(0+y)=f(0)+f(y) \iff f(y)=f(0)+f(y) \iff 0=f(0) $$ The same result is for when $y=0$  $x=-y:$ $$f(-y+y)=f(-y)+f(y) \iff f(0)=f(-y)+f(y) \iff 0=f(-y)+f(y)\iff \quad f(-y)=-f(y)$$ I want to extend the result of setting $x=-y$ to numbers other that $-1$, perhaps all real numbers or all rational numbers. I got a little help from reading other solutions on the next part:  Let $q=1+1+1+...+1$. Then  $$f(qx)=f((1+1+...+1)x)=f(x+x+...+x)=f(x)+f(x)+...+f(x)=qf(x)$$ I understood this part, but I don't understand why this helps me find all the functions that satisfy the requirement that $f(x+y)=f(x)+f(y)$, but here is how I went on:  Thus  $$f(qx)=qf(x)$$ and it should follow that $$f \bigg (\frac {1}{q} x\bigg)= \frac{1}{q}f(x)$$ where $q\not =0$, then it further follows that  $$f \bigg (\frac {p}{q} x\bigg)= \frac{p}{q}f(x)$$ where $\frac{p}{q}$ is rational, and lastly it further follows that  $$f (ax)= af(x)$$ where $a$ is real. Thus functions of the form $f(ax)$ where $a$ is real satisfies the requirement of  $f(x+y)=f(x)+f(y)$.  I don't know how much of what I did is correct\incorrect, and any help would be greatly appreciated. Also is there any way that I can say that functions of the form $f(ax)$ where $a$ is real are the only functions that satisfy the requirement of  $f(x+y)=f(x)+f(y)$? Or do other solutions exist?  Again, thanks a lot for any help! (Hints would be appreciated, I'll really try to understand the hints!) I am wondering if the mass density profile $\rho(\vec{r})$ has been characterized for atomic particles such as quarks and electrons. I am currently taking an intro class in quantum mechanics, and I have run this question by several professors. It is my understanding from the viewpoint of quantum physics a particle's position is given by a probability density function $\Psi(\vec{r},t)$. I also understand that when books quote the "radius" of an electron they are typically referring to some approximate range into which an electron is "likely" to fall, say, one standard deviation from the expectation value of its position or maybe $10^{-15}$ meters.  However it is my impression that, in this viewpoint, wherever the particle "is" or even whether or not the particle "had" any position to begin with (via the Bell Inequalities), it is assumed that if it were (somehow) found, it would be a point mass. This has been verified by my professors and GSIs. I am wondering if its really true.  If the particle was truly a point mass then wherever it is, it would presumably have an infinite mass density. Wouldn't that make electrons and quarks indistinguishable from very tiny black holes? Is there any practical difference between saying that subatomic particles are black holes and that they are point masses? I am aware of such problems as Hawking Radiation although at the scales of the Schwarchild radius of an electron (back of the envelope calculation yields $\tilde{}10^{-57}$ meters), would it really make any more sense to use quantum mechanics as opposed to general relativity?  If anyone knows of an upper bound on the volume over which an electron/quark/gluon/anything else is distributed I would be interested to know. A quick Google Search has yielded nothing but the "classical" electron radius, which is not what I am referring to.  Thanks in advance; look forward to the responses. Limits involving factorials $\lim_{N\to\infty} \frac{N!}{(N-k)!N^{k}}$ I am trying to calculate the following limit  $$\lim_{N\to\infty} \frac{N!}{(N-k)!N^{k}}$$ where $k$ can be any number between $0$ and $N$.  I thought of the following:  If I take the logarithm of the expression then I get:  $$\lim_{N\to\infty} \left(\log(N!)-\log((N-k)!)-k\log N\right)$$    Using the Stirling formula this can be approximated as:  $$\lim_{N\to\infty} \left(N\log(N)-(N-k)\log(N-k)-k\log N\right)$$  Now there are two cases: If $k$ is $N$, then the second term vanishes and the remaining terms cancel. If $k$ is smaller than $N$, then I can drop the $k$ inside the second logarithm and all the terms cancel. So the limit $$\lim_{N\to\infty} \log\left(\frac{N!}{(N-k)!N^{k}}\right)=0$$ Which means that: $$\lim_{N\to\infty} \frac{N!}{(N-k)!N^{k}}=1$$ I don't know if this is mathematically rigorous. Would like some help. Thanks What is a token? Please forgive such a basic question, but I did some Google searching and looked around on this site,  and while there are many threads and online resources that involve tokens, I can't seem to find a direct explanation of what a token is, specifically in the context of TeX, LaTeX, etc. Perhaps there is a good explanation of tokens in the TeXbook, but I don't have a copy of it.   gives me some idea:     A token is a string of characters, categorized according to the rules as a symbol (e.g., IDENTIFIER, NUMBER, COMMA).   but this is a bit too general for me to understand. Taking a guess based on the Wikipedia definition and , is it correct to say that a token is "a thing that has a catcode"? (which, even if true, does not seem entirely satisfactory as an answer: what characterizes the things that are given catcodes?)  Some illustrative examples, and non-examples, of tokens in the realm of TeX, LaTeX, etc. would be much appreciated. Which one is correct? (Maybe both are correct.)     He passed away on the morning of March 5.   Or      He passed away in the morning of March 5. Can I use Wi-Fi and cable (Ethernet) networking at same time? I am on an intranet that is not connected to the Internet and a Wi-Fi router with connection to the Internet. When I am connected to the wireless network, I can browse the Internet, but I cannot browse my LAN with my Ethernet cable connected.  How can I browse the Internet via my Wi-Fi router and at the same time browse my local LAN web application via my connected Ethernet cable? I have some files with TikZ graphics that have suddenly changed appearance since I upgraded to PGF-2.10.  Here is a minimal example:  \documentclass{minimal} \usepackage{tikz} \begin{document} \pgfversion\par \begin{tikzpicture} \draw[blue] plot[domain=-1:1] (\x,{\x^2 + \x^3}); \draw[xshift=4cm,blue] plot[domain=-1:1] (\x,{(\x)^2 + (\x)^3}); \end{tikzpicture} \end{document}   In PGF 2.00 I had this:    while in 2.10 I have this:    The one on the left in PGF 2.10 looks like the graph of y=x^3 but could also be y=x^3 + x|x|.   This has been  on sourceforge and  on the pgf-users mailing list.  Although I haven't checked myself yet, the bug remains open and some of the commenters below report that problems still exist in the CVS version.   Geoffrey Jones notes the bug is in the 2.00 CVS version:     Hi Matthew, not an answer, just an FYI   that's too large for the comment box.   My compile of your code through pgf   v2.00 produces the result you see with   v2.10, above:       I.e., the difference you observe seems   to pre-date v2.10.   There is a workaround by inserting parentheses.  But since I have so many old lecture files I'd rather find a solution that doesn't involve rewriting my TikZ code.  And I'm squeamish about using the CVS bleeding edge code even if it were fixed, since I don't want to introduce other problems.  Ideally I'd like a simple patch that I can stick at the preamble, or just fix one file in the PGF distro.