The English equivalent of the Danish word "lykkeridder" In Danish, a lykkeridder refers to a person who hopes to quickly achieve wealth or happiness through little labour. I am quite confident there is an English word for this -- it is on the tip of my tongue, but I can't recall it at the moment -- can someone enlighten me?
What's a good word for a person, or a state, when a person is constantly waiting for life to get good, but does nothing to actually make it happen I've written a song about people who constantly wait for something good to happen to them and am looking for a word to describe this state/paradox/condition.   The song is about people who expect life to get better for no reason, talk about moving to a bigger city or a different place as if that would magically make them rich, famous and happy, and keep talking about how they are going to do this or that, but never do it. I was under the impression that there's a psychological term for this state, but can't google it up, so I probably made it up.  This is essentially a song about myself, I keep talking about how I love to make music, or shoot videos, but I never actually produce anything, yet it's the only thing I can think of. When it comes to the moment to take action – I just stay at home watching tv shows.  I know it's a broad question and list of topics, but I guess I'm looking for a word that describes this state – of doing NOTHING.
How do I set the default program? MWE for externalize + automatic recompilation after tikz figure modification Word to describe the way cattle and buffalo consume fodder? How does one decide whether to use who or that? In the title example:   The correct pronoun is who. The correct pronoun is that. Either is correct. They're interchangeable. It depends on the context. When is a gerund supposed to be preceded by a possessive adjective/determiner? Pgfplots: How to fill area under a curve in a 3d plot similar to \closedcycle in 2d Contacting prospective advisors before graduate school applications - courtesy or annoyance? Last year, rather late during the MSc application cycle (after one round of funding was allocated - tardiness due to an abrupt change in my life situation), I emailed two professors from two schools that I most wanted to go to. I had perused their webpages and decided these professors had the specific research interests closest to my own.  My reason for contacting them before finishing my applications for their respective schools was that the applications included a large section for specific research questions that I wanted to investigate. So, I emailed these professors asking a bit about their work, because their webpages of course featured surface-level stuff.  One professor was very receptive and invited me to Skype with her so she could better explain herself and I could ask her questions. With her support, I was able to write an application that met her standards quite quickly.  The other professor told me - paraphrasing, of course - that he was too far into the process of choosing a new student for his research group to spend time helping me finish the application.  Let me be clear that the latter professor didn't say any harsh words and I have nothing bad to say about him. That said, the purpose of my question is to decide whether it's better for me as a prospective MSc student to email professors ahead of time like this. Was the latter's rejection likely due to my tardiness rather my emailing him? The more receptive professor made the point that she always wanted to have prospective MSc students write their own research proposals and get her approval before submitting the application. Is this common? If so, then I guess reaching out beforehand is a good thing to do, but earlier in the application cycle?  Thank you. Boot Error - ACPI PCC Probe Failed ACPI PCC Probe Failed is the error that pops up instead of the default GRUB loader after I updated my kernel to 3.19.  I searched around and couldn't find much information except that it is a known error and it is registered in kernel.org's launchpad . My question is, is anyone else facing it?, what are the impacts of this? And if only an update in the future can fix it, any suggestions or safety tips till then? Joining two rigged models I currently trying to join two rigged (and animated) models, which both has separate meshes, separate bones and separate animations. The first mesh is the horse while the second mesh is human:   So I am interested to join human model onto horse model to make a rider:   I think the best way would be to make horse as a parent and human as a child during the join process. In that case I would like that all horse animations (there are 3 of them - idle, walk and run) would remain as they are, while I can lose human animations, as I will need different ones for it. I was also thinking to make human root bone as a child of one horse back bones, that when horse moves up and down it would move entire human mesh together. It would be also good to keep the same weights that human bones would not start deforming horse mesh and vice versa.  I tried to use Ctrl+J by selecting various combinations between bones and meshes in object mode, however, I noticed that it messes up scaling and rotation of the child object. So I was wondering how it would be possible to properly join these models to create a rider? Boolean and Texts I have used Hexagon for years but find that it sometimes crashes when trying to do Boolean operations with texts.  I have been trying to make the two tables of the Ten Commandments and have been trying to use Blender.  I did it about two years ago in Blender and lost everything on my computer so I am trying again with the latest version of Blender and am having difficulty.  I start with a cube and then get it close to the dimensions I would think the Ten Commandments might look like. I then bevel the top using the bevel modifier. But in doing it this way I don't have any cross sub divisions on the face of the object. I have tried sub dividing first but then I find that I can't bevel the top arches as far as I would like. I still don't know how to add divisions after I bevel the top. Yes I am still a novice at this program.  In any case I did figure out how to make text and turn them into an object. When I make the texts I do all four of the commandments for one side and then the other six for the other side. So this makes allot of texts to cut out.  When doing the Boolean operation in Difference sometimes nothing happens. I am thinking I need to have some sub - divisions on the face of the sized cube after it is beveled but am not really sure if this is the problem.  This whole thing about normal's is still Greek to me. I have watched some videos on how to cut letters into a cube but for some reason it does not work this easy after I bevel the top of the cube. Or it could be that Blender does not like a bunch of texts all at once.  Could someone show me a video or give me some simple steps to make my project work? Exercise on local and global symmetry for a given Lagrangian density vi shows as vim When I type vi into a terminal it shows the following display:    Note that it says vim not vi, I'm sure I haven't downloaded vim yet and that this is actually vi not vim: for example the arrow keys print ABCD instead of moving. How do I determine the size of my array in C? How do I determine the size of my array in C?   That is, the number of elements the array can hold? I've recently been learning about differential equations, and my teacher has been giving some particularly difficult examples out for those of us who finish early. He gave us the following differential equation: $$x=y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^{2}$$ With the hint "Differentiate with respect to $y$, then let $\frac{dy}{dx}=p$". I've solved it after the lesson and would like someone to check whether or not I've done so correctly.    Following the hint, $$\frac{dx}{dy}=\frac{1}{p}=p+y\frac{dp}{dy}-2p\frac{dp}{dy}$$ Which rearranges to $$p-\frac{1}{p}+y\frac{dp}{dy}=2p\frac{dp}{dy}$$ Multiplying through by $\frac{dy}{dp}$, $$\left(p-\frac{1}{p}\right)\frac{dy}{dp}+y=2p$$ Rearranging into the form $\frac{dy}{dp}+f(p)y=g(p)$, and writing $p-\frac{1}{p}=\frac{p^{2}-1}{p}$ $$\frac{dy}{dp}+\frac{p}{p^{2}-1}y=\frac{2p^{2}}{p^{2}-1}$$ Our integrating factor is $$e^{\int\frac{p}{p^{2}-1}dp}=e^{\frac{1}{2}\ln(p^{2}-1)}=\sqrt{p^{2}-1}$$ So we have $$\sqrt{p^{2}-1}\frac{dy}{dp}+\frac{p}{\sqrt{p^{2}-1}}y=\frac{2p^{2}}{\sqrt{p^{2}-1}}$$ The left hand side is, by design, $\frac{d}{dp}(y\sqrt{p^{2}-1})$, and the right hand side can be integrated as follows: $$\int \frac{2p^{2}}{\sqrt{p^{2}-1}}dp=\int\frac{2(p^{2}-1)+2}{\sqrt{p^{2}-1}}dp=\int 2\sqrt{p^{2}-1}dp+\int \frac{2}{\sqrt{p^{2}-1}} dp$$ These can both be solved using the substitution $p=\cosh(u)$, and gives $$y\sqrt{p^{2}-1}=p\sqrt{p^{2}+1}-\cosh^{-1}(p)+\alpha$$ Where $\alpha$ is an arbitrary constant. Dividing, we finally have $$y=p+\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}$$  Here I was stuck for a long time, until I realised that we also have the equation we began with: $x=py-p^{2}$, which we can solve as a quadratic in $p$ to also get $$p=\frac{y\pm \sqrt{y^{2}-4x}}{2} \implies p^{2}=\frac{y^{2}+y^{2}-4x \pm 2y\sqrt{y^{2}-4x}}{4}$$  Thus, $$y=\frac{y\pm \sqrt{y^{2}-4x}}{2}+\frac{\cosh^{-1}\left(\frac{y\pm \sqrt{y^{2}-4x}}{2}\right)+\alpha}{\sqrt{\frac{y^{2}-2x \pm y\sqrt{y^{2}-4x}}{2}-1}}$$  This is as good as I can do - we have a relation between $y$ and $x$ - am I done? Is this the correct answer? Is there an easier way?     Thank you for your time.    Following the suggestion by @Valentin $$\frac{dy}{dp}=1+\frac{\frac{1}{\sqrt{p^{2}-1}}\sqrt{p^{2}-1}-p\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}=1+p\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}$$ Hence, $$dx=\frac{dy}{p}=dp\left(\frac{1}{p}+\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}\right)$$ What blueprints become available during story? As I progress, I realize that some blueprints open up throughout the story and don't have to be researched. Are these random or are they always the same and is there a list over those? Seems you can easily waste research point if you don't know that. Proof that every number ≥ $8$ can be represented by a sum of fives and threes. Can you check if my proof is right?  Theorem.  $\forall x\geq8, x$ can be represented by $5a + 3b$ where  $a,b \in \mathbb{N}$.  Base case(s): $x=8 = 3\cdot1 + 5\cdot1 \quad \checkmark\\ x=9 = 3\cdot3 + 5\cdot0 \quad \checkmark\\ x=10 = 3\cdot0 + 5\cdot2 \quad \checkmark$  Inductive step:  $n \in \mathbb{N}\\a_1 = 8, a_n = a_1 + (x-1)\cdot3\\ b_1 = 9, b_n = b_1 + (x-1)\cdot3 = a_1 +1 + (x-1) \cdot 3\\ c_1 = 10, c_n = c_1 + (x-1)\cdot3 = b_1 + 1 + (x-1) \cdot 3 = a_1 + 2 + (x-1) \cdot 3\\ \\ S = \{x\in\mathbb{N}: x \in a_{x} \lor x \in b_{x} \lor x \in c_{x}\}$  Basis stays true, because $8,9,10 \in S$  Lets assume that $x \in S$. That means $x \in a_{n} \lor x \in b_{n} \lor x \in c_{n}$.  If $x \in a_n$ then $x+1 \in b_x$,  If $x \in b_x$ then $x+1 \in c_x$,  If $x \in c_x$ then $x+1 \in a_x$.  I can't prove that but it's obvious. What do you think about this? What's the point of Hamiltonian mechanics? I've just finished a Classical Mechanics course, and looking back on it some things are not quite clear. In the first half we covered the Lagrangian formalism, which I thought was pretty cool. I specially appreciated the freedom you have when choosing coordinates, and the fact that you can basically ignore constraint forces. Of course, most simple situations you can solve using good old $F=ma$, but for more complicated stuff the whole formalism comes in pretty handy.  Then in the second half we switched to Hamiltonian mechanics, and that's where I began to lose sight of why we were doing things the way we were. I don't have any problem understanding the Hamiltonian, or Hamilton's equations, or the Hamilton-Jacobi equation, or what have you. My issue is that I don't understand why would someone bother developing all this to do the same things you did before but in a different way. In fact, in most cases you need to start with a Lagrangian and get the momenta from $p = \frac{\partial L}{\partial \dot{q}}$, and the Hamiltonian from $H = \sum \dot{q_i}p_i - L$. But if you already have the Lagrangian, why not just solve the Euler-Lagrange equations?  I guess maybe there are interesting uses of the Hamiltion formalism and we just didn't do a whole lot of examples (it was the harmonic oscillator the whole way, pretty much). I've also heard that it allows a somewhat smooth transition into quantum mechanics. We did work out a way to get Schrödinger's equation doing stuff with the action. But still something's not clicking.  My questions are the following: Why do people use the Hamiltonian formalism? Is it better for theoretical work? Are there problems that are more easily solved using Hamilton's mechanics instead of Lagrange's? What are some examples of that? "Error: Invalid signature" when configuring dual boot Ubuntu 14.04 and Windows 10 Dot Product Intuition I'm searching to develop the intuition (rather than memorization) in relating the two forms of a dot product (by an angle theta between the vectors and by the components of the vector ).      For example, suppose I have vector $\mathbf{a} = (a_1,a_2)$ and vector $\mathbf{b}=(b_1,b_2)$.  What's the physical or geometrical meaning that   $$a_1b_1 + a_2b_2 = |\mathbf{a}||\mathbf{b}|\cos(\theta)\;?$$  Why is multiplying $|\mathbf{b}|$ times $|\mathbf{a}|$ in direction of $\mathbf{b}$ the same as multiplying the first and second components of $\mathbf{a}$ and $\mathbf{b}$ and summing ?       I know this relationship comes out when we use the law of cosines to prove, but even then i cant get a intuition in this relationship.  This image clarifies my doubt:    Thanks how to make some border lines of a table thick and colored