How can i Change my java swing application title in Mac OS X title bar?
Java OS X Lion Set application name doesn't work
Case assignment rule is always executed in the test class Make Windows 10 stop installing driver software automatically Is there a string math evaluator in .NET? If I have a string with a valid math expression such as:  String s = "1 + 2 * 7";   Is there a built in library/function in .NET that will parse and evaluate that expression for me and return the result? In this case 15. There are many documentaries, forums, blogs and more dedicated to . I have been frantically searching for an answer to my question however none of my sources have clarity to the matter of hand. I would really love a clear explanation to: What exactly is Dark Matter? Please help me to have a clear understanding. How to solve $\frac 1{1.2.3}+\frac 1{2.3.4}+\frac 1{3.4.5}+...$ As my question says,   $$\frac 1{1.2.3}+\frac 1{2.3.4}+\frac 1{3.4.5}+...$$  As far as $\frac 1{1.2}+\frac 1{2.3}+\frac 1{3.4}+...$ is concerned, we can write it as $\frac 1{n-1} - \frac 1{n}$ What can we do here? Per User Named Credential and External Identity Users My question is almost similar to one mentioned in 'Auth Provider with per-user Named Credential' (). I am trying to authenticate and make a API call when a new or existing Salesforce community users try to log in. The Authentication is provided by 3rd party (Cognito). So, the requirement is :  User to be authenticated using the 3rd party system for login. After successful authentication and before community page is displayed, the API call is to be made to 3rd party system to get more details about the user.  In order to achieve this:  Created an Open ID Connect Auth. Provider Use the registration handler on the Auth. Provider to create/link a new/existing user. Use a login flow for API call. Use Per User Identity Type Named Credential to make API calls.  No, where I am having issue is once the user has authenticated successfully, how do I allow the user to make API call without authenticating it again. I am getting the error: An Apex error occurred: System.UnexpectedException: You don't have permission to view this data, enter credentials for the named credential in your personal settings. Or ask your Salesforce admin for help. I believe this error is due to the fact that the user is not listed under 'Authentication setting for External Systems'. My requirement doesn't allow me to authenticate the user again, so how do I achieve this using either APEX or any other way? In Ubuntu 13.04 there was an option to use the PPA by Relan in order to get ExFat functionality. Seeing that in Ubuntu this functionality is now available from the repositories without the PPA I tried this but failed to make it work.  So my question is how do I enable exfat support? My table is:     id home datetime player resource     1 10 04/03/2009 john 399   2 11 04/03/2009 juliet 244   5 12 04/03/2009 borat 555   3 10 03/03/2009 john 300   4 11 03/03/2009 juliet 200   6 12 03/03/2009 borat 500   7 13 24/12/2008 borat 600   8 13 01/01/2009 borat 700     I need to select each distinct home holding the maximum value of datetime. Result would be:     id home datetime player resource     1 10 04/03/2009 john 399   2 11 04/03/2009 juliet 244   5 12 04/03/2009 borat 555   8 13 01/01/2009 borat 700     I have tried: -- 1 ..by the MySQL manual:   SELECT DISTINCT   home,   id,   datetime AS dt,   player,   resource FROM topten t1 WHERE datetime = (SELECT   MAX(t2.datetime) FROM topten t2 GROUP BY home) GROUP BY datetime ORDER BY datetime DESC  Doesn't work. Result-set has 130 rows although database holds 187. Result includes some duplicates of home. -- 2 ..join  SELECT   s1.id,   s1.home,   s1.datetime,   s1.player,   s1.resource FROM topten s1 JOIN (SELECT   id,   MAX(datetime) AS dt FROM topten GROUP BY id) AS s2   ON s1.id = s2.id ORDER BY datetime   Nope. Gives all the records. -- 3 ..something exotic:   With various results. Unable to change payment information to NONE in iOS Appstore Ever since adding an iTunes gift card to my account, I'm getting verification requests like crazy. I can't even download free apps. It takes me to my payment info and when I try to select none, I continually get a message under the payment info that says "you must verify your payment info to continue"  I don't want a credit card on file and would really like the none option to work. My ID shows I still have iTunes credit. PCA on correlation or covariance? What are the main differences between performing principal component analysis (PCA) on the correlation matrix and on the covariance matrix? Do they give the same results? From a collection of short stories I read at least 30 years ago. :/  This is going to be incredibly vague, so apologies in advance.  A human and alien (blob/octopus-like...I think) from warring races are captured by an unidentified higher intelligence and put in an enclosure (more like a zoo habitat rather than a prison cell). They are separated by an invisible barrier and can see each other.  I think they were both trying to escape or trying to figure out how they could kill the other first. There was a bulldozer/loader type vehicle they could use at one point to build a shelter (or something).  I believe they were crafting makeshift weapons as well. There was no communication between the two as far as I remember.  The main tension in the story seemed to be was the race to get to the other first.  This has been stuck in my head forever. I know its a very common theme.  The short stories may have included Asimov. Cauchy-Schwarz inequality proof (but not the usual one) Before you downvote/vote-to-close, I am not asking for a proof of: $$\sum^n_{i=1}a_ib_i\le\sqrt{\sum_{i=1}^na_i^2}\sqrt{\sum^n_{i=1}b_i^2 } $$ Which is what EVERY link I've found assumes is the inequality (for a proof of that see:  )    I am reading a book that claims the Cauchy-Schwarz inequality is actually: $$\vert\langle x,y\rangle\vert\le\Vert x\Vert\Vert y\Vert$$ where $\Vert x\Vert :=\sqrt{\langle x,x\rangle}$  with the additional claim: equality holds $\iff\ x,y$ are linearly dependent  I cannot find a proof of this claim (only proofs for the dot product inner product). My question is: what is the simplest way to prove this (I define simplest to mean "using as few definitions outside of the inner product itself as possible" - so that's not including Gramm-Schmitt orthonormalisation, for example.)  I've just found some possible answers in the links on the right (annoyingly) but fortunately I have a second, albeit softer, question:  What inequalities that are common for say $\mathbb{R}^n$ are actually based on the inner product?   (In future I will check this site first, I was targeting lecture notes and such in my search)  Addendum: I want to use this on not-finite dimensional vector spaces. I've found one proof that relies on finite-ness. Ubuntu Software (Software Center) Crash - Ubuntu 16.04 When do we not put a comma before "because"? In what situation do you put a comma before "because"?   And, when do you not put a comma before "because"?   I've always assumed that you put a comma, because the clause introduced by "because" will always be independent, because it cannot stand on its own, because it wouldn't make sense on its own.   Is this the case? Could you think of an example where it doesn't take a comma? Is a general GRE score of X sufficient for entry to postgraduate degree Y? How can I move my /home directory to another partition if it's already part of the / partition? I have a wonderful running Maverick Meerkat install going on right now. I want to move my /home to another partition as per many suggestions.   How can I go about doing this without reinstalling the entire Ubuntu system? Can I simply create the partition with gparted and then change the mount point of /home in fstab to reflect it or is it a bit more difficult? What is the smallest positive integer $n$, such that there exist positive integers $a$ and $b$, with $b$ obtained from $a$ by a rearrangement of its digits, so that   $a – b = 11\dots 1$ (The number of '$1$'s equal to $n$)?   Is there any example that can satisfy the equation? Is it possible to run an x86 binary on an ARM processor? Evaluate $\lim_{n\to\infty} [{1\over kn}+{1\over k(n+1)}+{1\over k(n+2)}+\cdots+{1\over k(n+p-k)}]$ where $k<p$ There are four options. Which of them is correct?- (a)${p\over k}$ (b)${k\over p}$ (c)$\log({p\over k})$ (d)${\log p\over k}$ I somehow want to use the rule for evaluating this kind of sum using integration, i.e. $\int_{0}^{1} f(x) \ dx=\lim_{n\to\infty} {1\over n}\sum_{r=1}^{n}f({r\over n})$  But the given expression $\lim_{n\to\infty} [{1\over kn}+{1\over k(n+1)}+{1\over k(n+2)}+\cdots+{1\over k(n+p-k)}]={1\over k}\lim_{n\to\infty}{1\over n}\sum_{r=0}^{p-k}\frac{1}{1+{r\over n}}$, sum running from $0$ to $p-k$ instead of $n$. So, how to evaluate this limit? Thanks for assistance in advance. 2D collision detection Let's assume I'm using this character.  (source: ) How would you implement collision detection for it? Using a bounding box doesn't seem to be a good approximation, because the bird's shape is nowhere near a square. I was thinking of having a sort of quad tree data structure inside the object that represents portions of the image. Each leaf could be either false (in case it covers the white/transparent space outside the bird) or true (in case it represents an area of the bird i.e. beak, eye etc). Then somehow test the only obstacle in the scene for collision with the bird. But my problems in my approach are:  I don't know how to initialize the quad tree. Once the quad tree is initialized, I'm not sure how to traverse &amp; use it once the obstacle is within the coordinates of the image.  How would you do collision detection with non-squarish characters? LE: The other approach I've seen was to use multiple bounding boxes. For example I'd have one or a few bounding boxes for the beak, then a few of them for the hair or the tail. But it can get tedious. If this is a valid approach in my case, how would I generate those bounding boxes? I doubt I'd have to have them hardcoded in my program. LE2: I care about fairly precise collisions. I can't imagine how a single bounding box or circle can at least approximate decently that shape, so this approach won't work.