In the BibDesk wiki (BibDesk Tips and Tricks) there is some information on customizing BibDesk for biblatex, including a link to a TypeInfo file that can be used to create all of the biblatex fields automatically.  The has to be saved as ~/Library/Application Support/BibDesk/TypeInfo.plist.  Im using texmaker, bibDesk under osx 10.9.5. The folder doesn't exist so I created it. Didn't help.  Does anybody know where to copy the "TypeInfo.plist" file?   Alan Munn refers to this in that Question:
I’m currently in the process of reconsidering how I do bibliography management. It seems to be most sensible to use biblatex with biber. I’d like to know what tools are available on which I can build a solution that works for me. I’m interested in everything from full-blown bibliography management suites to pretty-printers for .bib files; both as traditional programs and online services.  Tools in question should fulfill the following criteria:   Unicode support. Support for the additional entry types and fields that biblatex supports. Support for biber’s improved cross-referencing.
CodeIgniter/PHP Active Record won't increment an integer What does = +_ mean in JavaScript Why do we use the object instead of the subject pronoun in constructions like "stupid me"? I'm trying to find out how come we say lucky me and stupid us rather than lucky I and stupid we. My understanding is that this is not a recent invention, but a relic from the distant past where it was a dative construction. However, I don't see the reason for using any case other than the nominative in this construction to begin with.        He went surfing. → Lucky him went surfing.   I got pulled over. → Stupid me got pulled over.   They bought a house. → Thoughtless them bought a house.      This is rather intriguing. After thinking about it for a while, I couldn't come up with any other language, Germanic or otherwise, where simply introducing an adjective would dictate a change of grammatical case. This appears to be an English peculiarity. How far can it be traced back? Can anyone comment on its background? How can I check the available version of a package in the repositories? How can I check the version of the available package in the Ubuntu repositories without installing it? The difficulty is that it should be cross platform. Windows 2000, XP, Vista, OSX, Linux, other unix variants. I am looking for a snippet of code that can accomplish this for all platforms, and a way to detect the platform.  Now, you should be aware of  that user.home does not work correctly, so please do not provide me of textbook answers, I can find these myself in the manuals. I have a model similar to below with a continuous outcome, continuous predictor and a categorical variable. the goal of this analysis is to make inferences on the group variable. The issue is that y is bounded by x in the form of y >= -x. When the model makes predictions you can see in the plot below that the predicted value is outside the possible range. See the last plot.   How can this be remedied? A custom loss function to penalize going beyond the bound and use gradient descent regression? Or is there some other way? Thank you.   x =   rep(seq(-10,10,1),each=5)  y =  rep(0,length(x) ) weights = sample( seq(1,1000,1) ,length(x)) groups = rep( letters[1:5], times =length(x)/5  )   ## the outcome y has the relationship y &gt;= -x for(i in 1:length(x)){   y[i] = sample(  seq(-x[i],-x[i]+ ifelse(x[i]&lt;0 ,                                            sample(seq(1,10,.1),1),                                           sample(seq(5,40,.1),1) ),1)                        ,1) }   dat =data.frame(x = x, y = y, weights = weights, group = groups) model = lm(dat = dat, y ~ x + group + group * x, weights = weights) dat$FIT= predict(model, dat, allow.new.levels = TRUE)  library(ggplot2) ggplot(data = dat, aes(x = x, y = y, color = group))+geom_point( aes(size = weights))+ geom_line(data = dat, aes(x = x, y = FIT, color = group))    #data for new predictions newdata = data.frame( x = seq(-100,100,1) ,                        group = rep("c", length( seq(-100,100,1) )),                       bound = -seq(-100,100,1)                       )  ## when pridicting new data the outcome y has the relationship y &gt;= -x ## but the predicted vales are below the bound. See how red line is below blue to the left of zero new_preds  = predict(model, newdata, allow.new.levels = TRUE) ggplot(data = newdata, aes(x = x, y = new_preds, color = group))+geom_line()  +   geom_line(data = newdata, aes(x = x, y = bound, color = "green")) check if "it's a number" function in Oracle test fails because of TestMethod do not support getContent call My below test fails because of Run Failures: TestRangehoodObjectionSheet.RangehoodObjectionSheetTest Methods defined as TestMethod do not support getContent call. I added the Blob thinking it might resolve but it doesn't work. What could be the problem here? Have I used the Blob correctly?  @isTest public with sharing class TestRangehoodObjectionSheet {      static testMethod void RangehoodObjectionSheetTest() {             String Build                = NBATRecords.getIdFromDeveloperName('Plan_Reviews__c','Building_Plan_Review');             Account a = new Account(Name = 'Test1' ,Type = 'Lead Acct',BillingState = 'CA',Industry = 'Business Services');             insert a;              Establishment_Record__c er = new Establishment_Record__c(Account__c=a.Id ,Project_Name__c = 'TestPreet',PremiseStreet__c='Manhattan ave',PremiseBorough__c='Manhattan',PremiseState__c='NY',PremisePostalCode__c='11102',Referral_Source__c='311',Data_Source__c='CAMIS',Phase__c='Initiate',Business_Status__c='Open' );             insert er;              Application__c ap = new Application__c(Establishment_Record__c=er.Id,Job_Number__c='222222222');             insert ap;              Plan_Reviews__c pr = new Plan_Reviews__c(Application__c =ap.Id ,DOB_Job_Number__c='222222222',RecordTypeId =Build,Objections__c='Pipe size not indicated');             insert pr;             PageReference pageRef = Page.RangehoodObjectionSheet;              Test.setCurrentPage(pageRef);               Blob content = Test.isRunningTest() ? Blob.valueOf('UNIT.TEST') : pageRef.getContent();              Plan_Reviews__c              p   = new Plan_Reviews__c(id=pr.Id);                ApexPages.StandardController sc  = new ApexPages.standardController(p);             System.assertNotEquals(sc,null);             RangehoodObjectionExtention    ext = new RangehoodObjectionExtention(sc);              ext.Savecopy();       } }   Actual class for the test:  public with sharing class RangehoodObjectionExtention { private final ApexPages.StandardController controller; private final Id                           planreviewID; public String imageURL{get;set;} public List&lt;String&gt; objlist {get;set;}  public string currentdate {get; set;}    public RangehoodObjectionExtention (ApexPages.StandardController stdController) {         controller   = stdController;         planreviewID = (Id)controller.getRecord().get('Id');         Date dt = Date.today();         currentdate = dt.format();         list&lt;Plan_Reviews__c&gt; prlist = [select FDNY_Objections__c from Plan_Reviews__c where id = :planreviewID limit 1];         string obj;         for(Plan_Reviews__c pr: prlist){            obj = pr.FDNY_Objections__c;             }         if(obj!=null){            list&lt;string&gt; objections = obj.split(';');             integer i =1;            objlist = new list&lt;string&gt;();            for(string o : objections){               objlist.add(i++ +'. '+o);            }         }         imageURL='/servlet/servlet.FileDownload?file=';         List&lt; document &gt; documentList=[select name from document where Name='RangehoodObjectionLogo'];         if(documentList.size()&gt;0)         {            imageURL=imageURL+documentList[0].id;         }       }          public PageReference Savecopy(){       PageReference p = Page.RangehoodObjectionSheet;       blob b = ApexPages.currentPage().getContentAsPDF ();       Attachment a = new Attachment();       a.body = b;       a.name = 'Rangehood Objection sheet Issued '+currentdate+'.pdf';       a.parentid = planreviewID;        insert a;        PageReference pref = new PageReference('/'+planreviewID);       return pref;    } } No homomorphism from $Z_{16}\oplus Z_{2}$ onto $Z_{4}\oplus Z_{4}$. For some reasons, homomorphism is a very hard area for me to make improvements. I've been hitting the brick wall for almost 2 hours.   Prove that no homomorphism exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$.  Assume a homomorphism $\phi$ exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$. Then $\phi$ is an isomorphism.  Note that the $\left | \ker \phi \right |=2$ The $\ker \phi$ is also a normal subgroup for $Z_{16}\oplus Z_{2}$. We want possible normal subgroup of order 2. I.e., 2 elements in each normal subgroups.  By Lagrange's theorem, the order of each element in a group divides the order of a group so the possible order of the elements are 1 or 2. If the elements has order 1, then the Ker \phi cannot have 2 elements.  Hence, $\ker\phi$ = $\left \{ (0,0),(8,0) \right \}$ ,$\left \{ (0,0),(0,8) \right \}$,$\left \{ (0,0),(8,1) \right \}$, possibly.  By the First isomorphism theorem:  $\phi: Z_{16}\oplus Z_2 \rightarrow Z_2\oplus Z_2$  $\Psi: G/\ker \phi \rightarrow \phi\left ( Z_16\oplus Z_2 \right )=Z_2\oplus Z_2$  $g\ker \phi  \mapsto \phi(g)=\Psi(g\ker \phi)$  Any help is appreciated.  Edit: I know that $(2,0)$ has order $8$ in $Z_{16}\oplus Z_{2}$ but any elements in $Z_{4}\oplus Z_{4}$ does not have order $8$ which would have solved the question at the outset. But I would like a different route. Who took care of Naruto when he was young? In episode 350, we see that the Third Hokage took on to take care of Naruto. But towards the beginning of the anime Naruto is seen to be living alone.   Did Naruto live with the Third? If so, why did he leave there? Choosing equipment options at character creation So if I'm correct, when you chose your starter gear and wealth you can either choose to roll some dice depending on your class and buy everything from that or you can choose the starter gear from your class and add the background items (including some gold). Here are my 2 questions:   If I choose the gear from my class, do I have to take all the equipment from only one of the (a) or (b) choices, or can I freely mix-and-match choices from the two sets? Can you spend the extra gold you get from your background on extra equipment (e.g. some soap and some fishing equipment) or do you have to start with that gold in its pouch?   The reason I ask is that I have recently created a level 1 wood elf druid and I mixed-and-matched the gear from the class, so that I chose a shortbow as a "(b) simple weapon" for the first point and a "(a) Scimitar" at the second point. My DM wasn't so happy about this, as he can't understand why I'm mixing (a) and (b).  He also wasn't so happy with me spending my 10gp from my outlander background. I decided that my druid wouldn't lay so much worth in 10gp compared to a bunch of stuff I got for that (e.g. some soap and some fishing equipment and some other useful things) and he seems to think that I'm supposed to just start with the gold instead. How to change the background color only for the current page? Can we denied permission for synchronized data extension in Salesforce Marketing Cloud, My requirement is, I don't want to allow users to access synchronized data extension either in contact builder or in Subscribers. Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Apparently $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$.  How? What's the proof? Or maybe it is self apparent just looking at the above?   PS: This problem is known as "The sum of the first $n$ positive integers". Direct proof that nilpotent matrix has zero trace $\mathbb{Q}^{\times}$ is the group of rational number without $0$ under multiplication, and $\mathbb{Z}^{n}$ is the free abelian group of rank $n$. Show that $\mathbb{Q}^{\times}$ not isomorphic to $\mathbb{Z}^{n}$. I tried to assume that there is an isomorphism and to get a contradiction.. didn't succeed Difference between pre-increment and post-increment in a loop? Is there a difference in ++i and i++ in a for loop? Is it simply a syntax thing? Brief Description:  I need my Ethernet card (en0) to see all traffic under 192.168.2.xxx.  However I also need to use WiFi card (en1) for all other traffic.  OS: OS X Lion    I am using a separate network location to deal with this specific use (since this is a for a robotics project that communicates via Ethernet). Ethernet (en0) had to be set up as a static IP (192.168.2.10) per the requirements for the device I am connecting to. All connections on the Ethernet card (en0) should be 192.168.2.xxx as specified above. WiFi (en1) is set up as DHCP to a router since their is no point in assigning it as a static for general purpose traffic. Ethernet (en0) is set up as priority over WiFi (en1).  I would also like to keep this under the network location preferences if at all possible since I do not require this feature to be on all the time.  I would really appreciate your help on this. No one I have talked to knows how to solve this. Find a bijection from $\mathbb R \to  \mathbb R-\mathbb N$. I want to set my function up such that all natural numbers get mapped to $n+.1$ and reals of the form $n+.1$ to $2n+.1$ is this correct? The sum of powers of two and two's complement – is there a deeper meaning behind this? Probably everyone has once come across the following "theorem" with corresponding "proof": $$\sum_{n=0}^\infty 2^n = -1$$ Proof: $\sum_{n=0}^\infty q^n = 1/(1-q)$. Insert $q=2$ to get the result.  Of course the "proof" neglects the condition on $q$ for this formula, and the sum really diverges. However I now noticed an interesting fact:  If you use two's complement to represent negative numbers on computers, $-1$ is represented by all bits set. Also, sign extending to a larger number of bits (that is, getting the same number in two's complement representation on more bits) works by copying the left-most bit (also known as sign bit) into the additional bits on the left.  Now imagine that formally you sign-extend the number $-1$ to infinitely many bits. What you get is an infinite-to-the-left string of $1$s. Which, using the normal base-2 formula $n = \sum_k b_k 2^k$ (where $b_k$ is the bit k positions from the right, i.e. $b_0$ is the rightmost bit), that infinite string of $1$s translates into exactly the sum above! So in some sense we have an independent re-derivation of that equation.  Now my question is: Is there something deeper behind this? Somehow I cannot imagine it is just coincidential.