Past participle - learned or learnt? Which is it? Personal preference, I dislike 'learnt', but it's correct in UK English usage - is that right?
When do you use “learnt” and when “learned”? Is learnt UK English and learned US? Is it that simple?  I’m used to using learnt, but my US spellchecker says it is wrong.
Is the function $\mbox{tr}(XAX')$ convex? Why does my regular expression work in X but not in Y? Using Custom Labels in a Class I have the below If statements using some hard coded strings and I want to make them use as custom labels. I have created custom labels with the same String names as given below. So how can I make these statements to use the Custom Labels I created ?  if(monitoringFrequency == 'daily'){             return dateValue.addDays(1);         }else if(monitoringFrequency == 'weekly'){             return dateValue.addDays(7);         }else if(monitoringFrequency == 'monthly'){             return dateValue.addMonths(1); Invisible software updates list in Software Updater Today I received an update notification for three updates but update manager doesn't say what updates are they.    Does anyone have a clue what this is about? How am I to check what is to be updated? Saving Working Files as PSD or TIFF? I was reading a few articles about how people saved their working files and it got me thinking about my own method. I used to save everything as a PSD, but recently I have started saving my working files as TIFF's. I use Windows, so I don't know if it's different for Apple users but with a PSD, all I can see is that "PSD" icon. But a TIFF file shows up as a normal image (that I can edit), so I found it really great when I have a lot of working files for one project and could just quickly preview it without opening the program.  That being said, I am no expert in which method is better. And I know a lot of people here are pretty seasoned users in all things Adobe, so I was wondering what your two cent on the whole topic is?   Is there really any difference to saving it as a PSD or TIFF? Is there something I should specifically know about saving my files as a TIFF rather than a PSD? If $U$ is unitary operator then spectrum $\sigma(U)$ is contained inside the unit circle In a Hilbert space, let $U$ be a continuous operator which it unitary. Prove $\sigma (U)\subseteq \Bbb{S}^1$.   It is important for me to know how I am doing, and I didn't come by a clear explanation so it would be appreciated to have your correcting and guiding.  Attempt: $U^{*}U=UU^{*}=I$ and therefore $U^{-1}=U$. I have $$ \| U x \|^2 = \langle U x, U x \rangle = \langle U^* U x, x \rangle = \langle x, x \rangle = \|x\|^2 $$ and the same goes for $U^*$ and therefore $\| U \| = \| U^{-1} \| = 1$.   Looking at $U-\lambda$ I can take $$ -\lambda U(U^{-1}-{1\over \lambda}I) =U-\lambda I. $$ Clearly, for $\lambda \ne 0$, $-\lambda U$ is defined and is a linear operator. $(U^{-1}-{1\over \lambda}I)$ is invertible if $|\lambda|\ge 1$ and therefore $\sigma (A^{-1})=\sigma (A)\subseteq \{|z|\le1\}$. But from the RHS, $U-\lambda I$ is invertible if $|\lambda|\le 1$ and therefore $\sigma(A)\subseteq \Bbb{S}^1$. I have noticed my claims tend to have holes in them many times which I cannot spot. If it is substantially true, how can I make sure it is completely formal? Object on curve wont stay flat! tkz-graph: changing individual vertex type I'm using these settings for most vertices:  \GraphInit[vstyle=Classic] \tikzset{VertexStyle/.append style = {minimum size = 5pt, inner sep = 0pt}}   But I'd like to have one or two vertices of my graph be hollow with a thick outer edge. I'm very confused about what options can be used in \Vertex, which ones can go in \SetUpVertex, for which I have to use \tikzset, etc. What's the best option here? I have 2 tables. One (domains) has domain ids, and domain names (dom_id, dom_url).  the other contains actual data, 2 of which columns require a TO and FROM domain names. So I have 2 columns rev_dom_from and rev_dom_for, both of which store the domain name id, from the domains table.  Simple.  Now I need to actually display both domain names on the webpage. I know how to display one or the other, via the LEFT JOIN domains ON reviews.rev_dom_for = domains.dom_url query, and then you echo out the dom_url, which would echo out the domain name in the rev_dom_for column.  But how would I make it echo out the 2nd domain name, in the dom_rev_from column? These wonderful icon overlays that give you useful information in the Windows Explorer are not showing up.   It's the most annoying with Dropbox (I already read ). I just tried reinstalling Dropbox - no help. I have this problem also with the  so I don't think that it is a dropbox problem specifically; and something is wrong in my OS. The OS is actually a pretty fresh clean-install...  OS: Win8 Pro 64-bit. My Visual Studio (2008) Editor has stopped to underline Errors (this nifty wavy red lines). I can't really tell when, but it can be related to the installation of .Net Framework 3.5 SP 1 or the MVC Beta (which I guess is unlikely). Furthermore have I installed and uninstalled both CodeRush and Resharper for evaluation purposes (decided not to keep either one of them).  Does anyone know the problem and how to restore this functionality again? Is the following layout possible with the subfigure package? I am using 301 redirects for all URLs.  Should I change sitemap URLs now?  The sitemap currently contains the following URLs:   example.com/cat/1.html example.com/cat/2.html example.com/cat/3.html   Should I change the sitemap's URLs to a new destination?   example.com/cat/year/1.html example.com/cat/year/2.html example.com/cat/year/3.html It says that Cognito Forms can send email notifications when a form is submitted. Where would I find those email notification settings?   I would like for my website clients to be the recipients of those email notifications. Do they have to have an account or a verified email address?   Also, is the form data included in the email notification? Also, is there an advantage to using the WordPress Plugin vs the regular embed code? How does this wiring work? I have a Trane XV90 and XL 14i. The wiring looks a little strange as in the picture bellow.                    _____________                   |           |   BK W2  W1   R  G  Y  B/C  YLO O     |___||   |__|  ||   |          |   |     ||   |_____>|----------|         |   |     ||_________>|other wire|         |   |     |           ------------         V   V     V     ---------------     |thermostat   |     ---------------  The R and G are jumpered together and the G wire is connected on the board  but is actually not connecting to the thermostat on the other end. So people told me that means the furnace fan is always on! (In my diagram above, I switched the location of Y and B/C to make diagram easier to draw. You can refer to the photo added below to see the details) My questions are the following  is the jumper between O and Y doing anything? I don't think I have heat pump I need to install Honeywell wifi thermostat and I am thinking of doing either one of the following:   remove jumper between R and G, and use G to connect B/C and then to thermostat C and put jumper between G and Y  leave jumper between R and G, and use G to connect B/C and then to thermostat C and put jumper between G and Y  remove jumper between R and G, connect G on control board to thermostat G, and finally have the spare wire to connect B/C and then thermostat C  leave jumper between R and G, does not connect G to thermostat,  and just use a spare wire to connect B/C on board to thermostat C   Can anyone comment on those plans? Thanks a lot! Cleaning my Macbook Pro Retina leaves greasy 'streaks' across the screen Let $P$ be a probability measure on $(\Omega, \mathscr{F})$, If $a,b,\in\mathscr{F}$ and $A\subset B$, then $P(B)-P(A)=P(B\setminus A)\ge 0$.  So far I have:  $A\subset B \implies B\setminus A = B\cap A^c = \emptyset$ as $A$ and $B^c$ are disjoint. Since $A\cup (B\setminus A)=A\cup (B\cap A^c)=(A\cup B)\cap (A\cup A^c)=B\cap\Omega = B$  Then for $A,B\in \mathscr{F}\implies P(A\cup (B\setminus A))=P(A)+P(B\setminus A) = P(B)$ by countable additivity.  Rearranging, $P(B)-P(A)=P(B\setminus A) \ge 0$. After looking into the different OAuth types and reading , I'm still confused on the specific pros and cons of user-agent vs web server flows.  The best explanation I can find says that user-agent tends to be used for mobile but it seems that they can both be used for desktop.   What makes it more appropriate to use one over the other in certain scenarios? The new 17.10 only has 64 bit ISOs. However 32 bit is still supported, there just are no live ISOs for it.  I have a 32 bit machine with WIFI, and I need to install from a USB disk. I tried the "mini.iso" but it doesn't have WIFI support.  What options do I have?  (PS I don't want another distro, and I can't use ethernet for installation, must be WIFI). Prove that the dihedral group $D_4$ can not be written as a direct product of two groups I like to know why the dihedral group $D_4$ can't be written as a direct product of two groups. It is a school assignment that I've been trying to solve all day and now I'm more confused then ever, even thinking that the teacher might have missed writing out that he means normal subgroups.  On another thread it was stated (as the answer to this question) that the direct product of two abelian groups is again abelian. If we consider the direct product of abelian subgroups $H$,$K\in G$ where $HK=G$ (for all $g \in G$, $g=hk$ $h \in H$, $k \in K$.) I can't understand why this would imply $g=kh$? It is not stated anywhere that $H$,$K$ has to be normal! But if this implication is correct I do understand why $D_4$ (that is non-abelian) can't be written as a direct product of two groups. But if it's not, as I suspect, I need some help!  We know that all groups of order 4 and 2 are abelian, (since $4=p^2$), but only 4 of the subgroups of $D_4$ are normal:  Therefor I can easily show that $D_4$ can't be a direct product of normal subgroups: The only normal subgroups of $D_4$ is the three subgroups of order 4, (index 2 theorem): $\{e,a^2,b,a^2b\}$, $\langle a\rangle$, $\{e,a^2,ab,a^3b\}$ and the center of $D_4=\{e,a^2\}$ We can see that these are not disjoint. So $D_4$ can't be a direct product of normal subgroups. The reason for this being that the center is non-trivial. But why can't $D_4$ be a direct product of any two groups?  If we write the elements of $D_4$ as generated by $a$ and $b$, $a^4=e$, $b^2=e$, $ba=a^3b$ why isn't $D_4=\langle b\rangle\langle a\rangle$ ? I calculated the products of the elements of these two groups according to the rule given above and ended up with $D_4$, and also $\langle a\rangle$, $\langle b\rangle$ is disjoint...? Why is this wrong? Very thankful for an answer!