How to pass a globbing pattern as parameter to a function in bash?
Why does my shell script choke on whitespace or other special characters?
Handling jQuery Component Collision In plugin development, what is the best practice to prevent jQuery component collision on the front end?  For instance, let's say I include the jQuery Apprise dialog in a plugin that loads this on the front-end for something, yet another plugin may do the same. Because this gets loaded and declared twice, or perhaps one is forked and customized while mine is not, we get Javascript errors on the frontend (I assume).  (Note that I'm doing the best practice of using the wp_register_script() and wp_enqueue_script() strategy via a wp_head() action event in order to load a jQuery component on the frontend.) Continuous function from $(0, 1]$ onto $(0, 1)$? Extreme of $\cos A\cos B\cos C$ in a triangle without calculus. If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.   I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing  helps.  On this topic we learned also about Cauchy inequality, but I have no experience with it.  The answer according to Mathematica is when $A=B=C=60$.  Any ideas? How to work around Eclipse Luna/GTK bug 339174 (change to not use oxygen) The bug is described here:  For me it fails with:  java: /build/buildd/oxygen-gtk3-1.0.2/src/animations/oxygencomboboxdata.cpp:87: void Oxygen::ComboBoxData::setButton(GtkWidget*): Assertion `!_button._widget' failed.   Problem is if I go to GTK+ Apperance in Control Module and change to anything other than oxygen Eclipse will look totally awful.  Regardless what theme I use (tried QtCurve and Raleigh) it has the same look with huge padding around all controls etc.  Is it possible to have a sensible look without oxygen GTK?  I have kubuntu-desktop installed over the stock LTS distro.   $dpkg -l libgtk* | grep -e '^i' | grep -e 'libgtk-*[0-9]'     ii  libgtk-3-0                             3.4.2-0ubuntu0.8                          GTK+ graphical user interface library     ii  libgtk-3-bin                           3.4.2-0ubuntu0.8                          programs for the GTK+ graphical user interface library     ii  libgtk-3-common                        3.4.2-0ubuntu0.8                          common files for the GTK+ graphical user interface library     ii  libgtk2.0-0                            2.24.10-0ubuntu6.1                        GTK+ graphical user interface library     ii  libgtk2.0-bin                          2.24.10-0ubuntu6.1                        programs for the GTK+ graphical user interface library     ii  libgtk2.0-common                       2.24.10-0ubuntu6.1                        common files for the GTK+ graphical user interface library It is quite easy to calculate the total number of functions from a set $X$ with $m$ elements to a set $Y$ with $n$ elements ($n^{m}$), and also the total number of injective functions ($n^{\underline{m}}$, denoting the falling factorial). But I am thinking about how to calculate the total number of surjective functions $f\colon X \twoheadrightarrow Y $.  The way I thought of doing this is as follows: firstly, since all $n$ elements of the codomain $Y$ need to be mapped to, you choose any $n$ elements from the $m$ elements of the set $X$ to be mapped one-to-one with the $n$ elements of $Y$. This results in $n!$ possible pairings. But the number of ways of choosing $n$ elements from $m$ elements is $\frac{m!}{(m-n)!\,n!}$, so the total number of ways of matching $n$ elements in $X$ to be one-to-one with the $n$ elements of $Y$ is $\frac{m!}{(m-n)!\,n!} \times n! = \frac{m!}{(m-n)!}$.  Now we have 'covered' the codomain $Y$ with $n$ elements from $X$, the remaining unpaired $m-n$ elements from $X$ can be mapped to any of the elements of $Y$, so there are $n^{m-n}$ ways of doing this. Therefore I think that the total number of surjective functions should be $\frac{m!}{(m-n)!} \, n^{m-n}$.  Is this anything like correct or have I made a major mistake here? Is it possible to overload the null-coalescing operator for a class in C#?    Say for example I want to return a default value if an instance is null and return the instance if it's not. The code would look like something like this:     return instance ?? new MyClass("Default");     But what if I would like to use the null-coalescing operator to also check if the MyClass.MyValue is set? When can lands be tapped? A dollar amount $\${-}67.9{-}$ is divisible by 72. What is this number? A client buys 72 turkeys. On the receipt, 2 digits are missing, the first and the last. So, the total price is $\${-}67.9{-}$. Find a way to make the price of each turkey round at the second digit without trial-and-error method.  So, the answer is obviously $\$5.11$/turkey, so $\$367.92$ total. I found that with with 2-3 trial-and-error.  I bypassed his problem by using python and made a script that tries every combination and assures it is divisible by 0.01.   Please help me find a mathematical way of solving this problem. What Canon lens would one use for Sartorialist-like street photography? You probably know Scott Schuman and his blog (). What kind of lens (that is not expensive) would you recommend for Canon 550D to achieve similar effect (in terms of DoF etc.)? I'm a newbie so I'm not sure how to properly interpret those photos' EXIF info. My guess is, that a similar effect can be achieved using .   PS:   Update: Thank you guys for all the answers, it's much more clear for me right now. I think I'll try the Canon 50mm f/1.4 lens. integral of $\int_{0}^{1}\frac{\ln(x^{2}+1)}{x+1}dx$ using contour integration? I have an interest in contour integration. I am not that good at it, but I enjoy learning what I can about it.  Here is a version of a rational log integral rarely encountered.  $\displaystyle \int_{0}^{1}\frac{\ln(x^{2}+1)}{x+1}dx=\frac{3}{4}\ln^{2}(2)-\frac{{\pi}^{2}}{48}$  I can do this using real methods (via double integral and substitution). I can post my workings if anyone would be interested.  My question is, can this be evaluated using contour integration due to the limits being $[0,1]$ instead of $[0,\infty)$?.  Contours may not be the most efficient way to go about it, but what is the course of action when the limits are 0 to 1 instead of 0 to infinity?. Why is "night" spelled with "gh"? Were these other people Chronicoms? We know that Enoch is a Chronicom, which is to say a sort of alien temporal observer.   When Enoch and his agents came and kidnapped       Coulson’s team,   for the purpose of sending them where they wanted; were these other agents also Chronicoms? I’m inclined to think they must have been (would Enoch have shared their advanced technology and goals, etc. with mere mortals), but on the other hand only one pod was mentioned. How can I install Ubuntu without removing Windows? Is there any difference between "offense" and "offence"? Is there a way to avoid SSH typing delay? Verify my proof of the irrationality of $\sqrt{6}$. TestMethod do not support Web service callouts, test skipped How to show that every Boolean ring is commutative? Operator Exponential $e^A e^B = e^{A+B}$ This question comes from an exam in my functional analysis class.  Suppose $X$ is a Banach space, and $T \in B(X,X)$ is a bounded linear operator on $X$. For any non-negative integer $n$, let $$S_n=\sum_{k=0}^n \frac{1}{k!} T^k$$ where $T^k$ is the composition of $T$ with itself $k$ times and $T^0=I$.  We can show that for any integer $k&gt;0$, $\Vert T^k \Vert \le \Vert T \Vert ^k$. Then we can show that $S_n \in B(X,X)$ and there is some $S \in B(X,X)$ such that $S_n \to S$. We write $S = e^T$ for this operator.  Finally, We were asked to show that $e^T$ has an inverse, and that it is $e^{-T}$. My thought is to prove the following claim first: if $A,B \in B(X,X)$ and $AB = BA$, then $e^A  e^B = e^{A+B}$. If the claim is true, it follows that $e^T e^{-T}=I$.  The claim can be proven provided that the product series can be computed with the Cauchy rule.  $$e^Ae^B=\sum_{i=0}^{\infty}\frac{A^i}{i!}\sum_{j=0}^{\infty}\frac{B^j}{j!}=\sum_{k=0}^{\infty}\sum_{l=0}^{k}\frac{A^lB^{k-l}}{l!(k-l)!}$$ $$=\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{l=0}^{k}\frac{k!}{l!(k-l)!}A^lB^{k-l}= \sum_{k=0}^{\infty}\frac{1}{k!}(A+B)^k= e^{A+B}$$  But why can the product series be summed in the Cauchy way? I know for real-number series, by Cauchy's theorem, if $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ are absolutely convergent to $A$ and $B$, respectively, then we can add $a_i b_j$ in any way, and the resulting series will converge to $AB$. Does this proposition still hold for commutable operators? (It would be greatly appreciated if ideas of proof or reference is suggested.) Is there any way to set a system wide memory limit a process can use in Windows XP? I have a couple of unstable apps which do work ok for most of the time but can hit a bug which results in eating whole memory in a matter of seconds (or at least I suppose that's it). This results in a hard reset as Windows becomes totally unresponsive and I lose my work.  I would like to be able to do something like the /etc/limits on Linux - setting M90, for instance (to set 90% max memory for a single user to allocate). So the system gets the remaining 10% no matter what.