Error loading DLL in python, not a valid win32 application
Python Ctypes - loading dll throws OSError: [WinError 193] %1 is not a valid Win32 application
Instantiate TinyMCE via JS ( rather than PHP/wp_editor ) What is a good asymptotic for $f_n = f_{n-1}+\ln(f_{n-1})$? I've working LDAP authentication with the following setup   AuthName            "whatever"  AuthType            Basic  AuthBasicProvider   ldap  AuthLDAPUrl         "ldap://server/OU=SBSUsers,OU=Users,OU=MyBusiness,DC=company,DC=local?sAMAccountName?sub?(objectClass=*)"  Require ldap-group  CN=MySpecificGroup,OU=Security Groups,OU=MyBusiness,DC=company,DC=local   This works, however I've to put all users I want to authenticate into MySpecificGroup. But on LDAP server I've configured that MySpecificGroup also contains the group MyOtherGroup with another list of users.  But those users in MyOtherGroup are not authenticated, I've to manually add them all to MySpecificGroup and basically can't use the nested grouping. I'm using Windows SBS 2003.  Is there a way to configure Apache LDAP to do this? Or is there a problem with possible infinite recursion and thus not allowed? This applies to all d20 systems, but with an example from the from the Pathfinder  (emphasis mine):  Stonecunning: Dwarves receive a +2 bonus on Perception checks to potentially notice unusual stonework, such as traps and hidden doors located in stone walls or floors. They receive a check to notice such features whenever they pass within 10 feet of them, whether or not they are actively looking.  In this case, does the GM roll the check in secret or ask the player roll the check? If the GM asks the player to roll for the check, what stops situations like:  GM: (interrupting) &quot;Gimli, Roll a perception check.&quot; Dwarf player: &quot;Oh, it's a natural one, I guess I don't see anything.&quot; Human player: &quot;Okay, I'll roll a perception check for anything unusual in this room. Nice! 19. How about 19, does 19 work for you?&quot; GM: &quot;Gimli, everything looks kosher to you. Aragon, you notice a rock coloured tarp on the ground here, covering what looks like a pit trap.&quot;  In this case, by telling the dwarf to roll his &quot;free&quot; perception check, he's telegraphed to the players that there is something to notice. How can this be fixed, without bluffing checks for nothing in particular to throw off the signal to noise ratio of similar checks? Is it implied that Porpetina and Queenie Goldstein are Jewish? What are Harry, Ginny, Ron and Hermione's professions? After the end of the Second Wizarding War Harry Potter later married Ginny Weasley, and the couple had three children of their own: two sons James Sirius, Albus Severus, and one daughter Lily Luna. What are Harry and Ginny's professions? I heard that Harry became an Auror, but what about Ginny, Ron, and Hermione? A sum of series problem: $\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$ I have a question regarding the sum of this series: $$\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$$ My approach: I found that this sum is equal to: $$\sum_{n=3}^{2008}\frac{n}{(n-2)!+(n-1)!+(n)!}$$  I reduced it to : $$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}$$  Please suggest how to proceed further. I received an email saying documents are /were Concatenate multiple rows (I'm going to link to this Stackexchange post concerning Nullary Operations: )  In general an  is a function $f:S^n \to S$, where $n$ denotes the arity of the operation, and a  is a function $f:S^0 \to S$.  It is clear that $S^n=S \times S \times \cdots \times S$ denotes the Cartesian product of a set with itself ($n$ times), however $S^0$ makes little sense to me.  One might define $S^0=\emptyset$. At first glance, that seems like a good choice. However then a nullary operation makes no sense, since it must be a function $f: A \to B$ where $A$ and $B$ are two sets representing the domain and codomain of the function. And note a function is defined in terms of relation, where a relation is a subset of the Cartesian product $A \times B$. So when we think of $A=S^0=\emptyset$ and $B=S$, so that we have a nullary operation $f:\emptyset \to S$, what is a a subset of $\emptyset \times S$? What even is $\emptyset \times S$? By definition of the Cartesian product, \begin{equation} \emptyset \times S = \{(a,b):a \in \emptyset\textrm{ and }b \in S\}. \end{equation} But there exists no $a \in \emptyset$! Therefore there exist no ordered pairs in our Cartesian product! Therefore $\emptyset \times S = \emptyset !$. Remember a function is a relation which is a subset of the Cartesian product. Therefore our relation is the empty set, and therefore also $f=\emptyset$.    Edit: We got the definition sorted out in the comments. Thanks especially to Hayden and Christoff for their great help. In summary, we figured out that $S^0=\{()\}$, the set containing only one element, the "empty $n$-tuple". Then if taking $S=\{1,2,...,k\}$, then a nullary operation $f:S^0 \to S$ is a function, i.e. some subset of $S^0 \times S = \{()\} \times S$. Looking at $\{()\} \times S$, this equals $\{((),1),((),2),...,((),k)\}$. Then a function is a special subset of this, so that each element of the domain (and there is only one element, namely $()$) is paired with exactly one element of the codomain. Therefore a function would have the form $f=\{((),a)\}$ where $a$ is one of $1,2,...,k$. Thank you for your help. Command-line to switch between profiles in gnome-terminal I created a new profile for gnome-terminal and I can switch between &quot;Implicit&quot; profile and the new created profile as you can see in the below image:  Now I wonder how can I switch between profiles using a command-line/script or maybe using a shortcut for this.  Note: Solutions like: gnome-terminal --tab-with-profile=Implicit &amp;&amp; exit   are excluded because I prefer not to close and open another terminal or get another terminal window in this process of changing the profile. Links in newsletter preview link to YouTube The problem: The links at the bottom of the Stack&nbsp;Exchange newsletter link to a YouTube video instead of the right place on the website. I´m not sure if this is an oversight or a clever hack.  Screenshot: Does "+" use in String concatenation affect efficiency? I have a batch file which is in a directory and must be run from there as well because it updates files within this directory. This works perfectly fine, except when the user runs the batch file as administrator (required on Vista). Then the starting directory is C:\Windows\System32. Is there any way to still be able to know from which directory the batch file was run? I dont want the user to enter the directory manually. Watch Filesystem in Real Time on OS X and Ubuntu I'm looking for a CLI tool which will watch a directory and spit out the names of files that change in real time.   some_watch_command /path/to/some/folder | xargs some_callback   I'm aware of inotify (inotify-tools?) and it seems to be what I need, but I need something that is both Linux (in my case Ubuntu) and OSX compatible.  It doesn't need to be lightning fast, but it does need to trigger upon changes (within a second is reasonable). Also, I don't necessarily need the exact CLI program mentioned above.  If some underlying tech exists and is easily scriptable on both platforms that would be great too. How do you read C declarations? Why didn't Skynet kill Kyle Reese? How do I determine which graphics card I'm using? Closure of the set of degree $n$ polynomials with $n$ real roots A very beautiful (but maybe classic) exercise:   Consider the set $\Omega_n$ of polynomials of degree $n$ with real coefficients and $n$ distinct real roots.   What is the closure of $\Omega_n$ in $R_n[x]$?  ($R_n[x]$ denotes the set of polynomials with real coefficients of degree at most $n$)    The answer is the set of polynomials in $R_n[x]$ all of whose roots are real, plus the constant polynomials. Showing that this set is contained in the closure is not too hard, however the reverse implication is more difficult and I'm wondering what is the most elementary method to prove it (without talking about holomorphic functions for example). I have this HW where I have to calculate the $74$th derivative of $f(x)=\ln(1+x)\arctan(x)$ at $x=0$. And it made me think, maybe I can say (about $\arctan(x)$ at $x=0$) that there is no limit for the second derivative, therefore, there are no derivatives of degree grater then $2$. Am I right?